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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeJan 27th 2019
   • (edited Jan 27th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexHere I would like to chat, especially with Richard, about possible algorithms that could decide, given a unitary linear representation, which of its _Chern classes_ vanish (or given an orthogonal linear representation, which of its Pontryagin classes vanish). (This understood in the fairly evident sense, as described in more detail at _[[equivariant K-theory]]_ -- [here](https://ncatlab.org/nlab/show/equivariant+K-theory#ChernClassesOfLinearRepresentations).) There are at least two motivations for this: Purely mathematically, computing higher Chern classes of linear representations is clearly a fundamental question; but it is also an open problem, highlighted, way back, in the appendix of * Michael Atiyah, _Characters and cohomology of finite groups_, Publications Mathématiques de l’IHÉS, Volume 9 (1961) , p. 23-64 ([numdam](http://www.numdam.org/item/?id=PMIHES_1961__9__23_0)) Judging from the combined upvotes and lack of reply to [this MO question](https://mathoverflow.net/q/312396/381), this problem seems to remain as open as ever. Second, there is a deep physics problem associated with this, which I can say more about by private email, if desired. For the _first_ Chern class $c_1$ of a linear representation $V$, there is an explicit formula. I have spelled out this explicit expression for $c_1(V)$ in terms of the character $\chi_V$ [here](https://ncatlab.org/nlab/show/equivariant+K-theory#eq:FirstChernClassOfRepresentationInTermsOfTheCharacter). It comes out as: $$ c_1(V) = \chi_{\left(\wedge^n V\right)} \;\colon\; g \;\mapsto\; \underset{ { k_1,\cdots, k_n \in \mathbb{N} } \atop { \underoverset{\ell = 1}{n}{\sum} \ell k_\ell = n } }{\sum} \underoverset{ l = 1 }{ n }{\prod} \frac{ (-1)^{k_l + 1} }{ l^{k_l} k_l ! } \left(\chi_V(g^l)\right)^{k_l} $$ For example, for a representation of dimension $n = 2$ this reduces to $$ c_1(V) = \chi_{V \wedge V} \;\colon\; g \;\mapsto\; \frac{1}{2} \left( \left( \chi_V(g)\right)^2 - \chi_V(g^2) \right) $$ which one can also find in the standard literature (e.g [tom Dieck 09, p. 45](https://ncatlab.org/nlab/show/representation+theory#tomDieck09)). For the higher Chern classes no explicit such formulas seem to be known (going by Atiyah raising this as an open question, and the MO request yielding plenty of views but no reply). On the other hand, for that physics application which I have in mind, it would already be highly interesting just to know which of the higher Chern classes _vanish_, and which do not. For this simpler question, I think it should be easy to give an effective algorithm, by using the above formula for the first Chern class together with the [[splitting principle]]. I'll try to say this in more detail in the next comment. But I'd be happy if we could brainstorm a little and possibly find more or better insights.

   Here I would like to chat, especially with Richard, about possible algorithms that could decide, given a unitary linear representation, which of its Chern classes vanish (or given an orthogonal linear representation, which of its Pontryagin classes vanish).

   (This understood in the fairly evident sense, as described in more detail at equivariant K-theory – here.)

   There are at least two motivations for this:

   Purely mathematically, computing higher Chern classes of linear representations is clearly a fundamental question; but it is also an open problem, highlighted, way back, in the appendix of

   • Michael Atiyah, Characters and cohomology of finite groups, Publications Mathématiques de l’IHÉS, Volume 9 (1961) , p. 23-64 (numdam)

   Judging from the combined upvotes and lack of reply to this MO question, this problem seems to remain as open as ever.

   Second, there is a deep physics problem associated with this, which I can say more about by private email, if desired.

   For the first Chern class $c_1$ of a linear representation $V$, there is an explicit formula. I have spelled out this explicit expression for $c_1(V)$ in terms of the character $\chi_V$ here. It comes out as:

   $$c_1(V) = \chi_{\left(\wedge^n V\right)} \;\colon\; g \;\mapsto\; \underset{ { k_1,\cdots, k_n \in \mathbb{N} } \atop { \underoverset{\ell = 1}{n}{\sum} \ell k_\ell = n } }{\sum} \underoverset{ l = 1 }{ n }{\prod} \frac{ (-1)^{k_l + 1} }{ l^{k_l} k_l ! } \left(\chi_V(g^l)\right)^{k_l}$$

   For example, for a representation of dimension $n = 2$ this reduces to

   $$c_1(V) = \chi_{V \wedge V} \;\colon\; g \;\mapsto\; \frac{1}{2} \left( \left( \chi_V(g)\right)^2 - \chi_V(g^2) \right)$$

   which one can also find in the standard literature (e.g tom Dieck 09, p. 45).

   For the higher Chern classes no explicit such formulas seem to be known (going by Atiyah raising this as an open question, and the MO request yielding plenty of views but no reply).

   On the other hand, for that physics application which I have in mind, it would already be highly interesting just to know which of the higher Chern classes vanish, and which do not.

   For this simpler question, I think it should be easy to give an effective algorithm, by using the above formula for the first Chern class together with the splitting principle.

   I’ll try to say this in more detail in the next comment. But I’d be happy if we could brainstorm a little and possibly find more or better insights.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeJan 27th 2019
   • PermaLink
   Author: Urs
   Format: MarkdownItexSuddenly I found a bunch of relevant references that I had been missing before. Maybe the formula we need is already there: on the bottom of p. 4 in [Symonds 91](https://ncatlab.org/nlab/show/characteristic+class+of+a+linear+representation#Symonds91). But need to chase some references now...

   Suddenly I found a bunch of relevant references that I had been missing before. Maybe the formula we need is already there: on the bottom of p. 4 in Symonds 91. But need to chase some references now…

3. • CommentRowNumber3.
   • CommentAuthorRichard Williamson
   • CommentTimeJan 27th 2019
   • PermaLink
   Author: Richard Williamson
   Format: MarkdownItexUnfortunately I do not have access to the paper by Symonds, but this sounds fun! Though I am not really familiar with this material, I agree that the splitting principle together with the formula for the first Chern class already gives a formula for checking vanishing. To be able to compute the formula, one would need some algorithm for finding $k_{1}, \ldots, k_{n}$ given $l$, but this should be perfectly do-able.

   Unfortunately I do not have access to the paper by Symonds, but this sounds fun!

   Though I am not really familiar with this material, I agree that the splitting principle together with the formula for the first Chern class already gives a formula for checking vanishing. To be able to compute the formula, one would need some algorithm for finding $k_{1}, \ldots, k_{n}$ given $l$, but this should be perfectly do-able.

4. • CommentRowNumber4.
   • CommentAuthorDavidRoberts
   • CommentTimeJan 28th 2019
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItex@Richard go to <https://eudml.org/doc/140229> and click on 'Access to full text', it gives you an OA copy. A lot of European society-based journals end up on GDZ.

   @Richard

   go to https://eudml.org/doc/140229 and click on ’Access to full text’, it gives you an OA copy.

   A lot of European society-based journals end up on GDZ.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeJan 28th 2019
   • PermaLink
   Author: Urs
   Format: MarkdownItexThanks, David, have added that link to the entry.

   Thanks, David, have added that link to the entry.

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeJan 28th 2019
   • (edited Jan 28th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexLet me see: So the result of [Symonds 91](https://ncatlab.org/nlab/show/characteristic+class+of+a+linear+representation#Symonds91) seems to say the following: Let $G$ be a finite group and $V$ a finite-dimensional linear $G$-representation over the complex numbers, write $[V]$ for its isomorphism class regarded as an element in the representation ring $R_{\mathbb{C}}(G)$. Consider then any presentation of $[V]$ as a sum of virtual induced representations of 1-dimensional representations $W_i$, via the [[Brauer induction theorem]]: $$ [V] \;=\; \underset{ \mathclap{ {H_i \hookrightarrow G} \atop { W_i \in Rep(H_i) \,,\; dim(W_i) =1 } } }{\sum} n_i \left[ ind_{H_i}^G W_i \right] \,. $$ Then the main theorem of [Symonds 91](https://ncatlab.org/nlab/show/characteristic+class+of+a+linear+representation#Symonds91) (see introduction & p.4-5) says, if I am reading it correctly, that the Chern character of the representation $V$, in the sense of the formal sum of Chern classes, is $$ ch \left( V \right) \;=\; \underset{ \mathclap{ {H_i \hookrightarrow G} \atop { W_i \in Rep(H_i) \,,\; dim(W_i) =1 } } }{\prod} \mathcal{N}_{H_i}^G ch\left(W_i\right)^{\alpha(W_i)} $$ where 1. the "multiplicative transfer" maps $\mathcal{N}_J^G$ are to be found in [Evens 63](https://ncatlab.org/nlab/show/characteristic+class+of+a+linear+representation#Evens63) 1. the $\alpha(W_i)$-s are the Euler characteristics of certain CW-complexes, described on p. 3 of [Symonds 91](https://ncatlab.org/nlab/show/characteristic+class+of+a+linear+representation#Symonds91). So for detailed computation we next need those $\mathcal{N}$ and $\alpha$. Currently I have no feeling for how tractable these are. But I am thinking now that a special case of particular interest to me can now immediately be treated with this formula. But please give me a sanity check here, because my conclusion seems a little funny: Namely suppose we happen to take $V$ to be a virtual permutation representation (hence to be in the image of the canonical morphism $A(G) \overset{\beta}{\to} R_{\mathbb{C}} G$ from the [[Burnside ring]]). Since the basic permutation representations $\mathbb{C}[G/H]$ are induced from the _trivial_ 1-dimensional $H$-representation $$ \mathbb{C}[G/H] \; \simeq \; ind_H^G triv_{1d} $$ it should follow that in this case the Chern characters on the right hand side of the above formula are all trivial $$ ch(W_i) =ch(triv_{1d}) = 1 \in H^0(B G) \hookrightarrow \prod_{n} H^{2 n}(B G) $$ ?! Plugging this into Symonds' formula above would imply that _all_ Chern classes $c_{k}$, $k \geq 1$, of virtual permutation representations just vanish. ?! Maybe I am making a mistake here, please let me know if you see where I am going astray. If this is true, it would be... interesting.

   Let me see: So the result of Symonds 91 seems to say the following:

   Let $G$ be a finite group and $V$ a finite-dimensional linear $G$-representation over the complex numbers, write $[V]$ for its isomorphism class regarded as an element in the representation ring $R_{\mathbb{C}}(G)$.

   Consider then any presentation of $[V]$ as a sum of virtual induced representations of 1-dimensional representations $W_i$, via the Brauer induction theorem:

   $$[V] \;=\; \underset{ \mathclap{ {H_i \hookrightarrow G} \atop { W_i \in Rep(H_i) \,,\; dim(W_i) =1 } } }{\sum} n_i \left[ ind_{H_i}^G W_i \right] \,.$$

   Then the main theorem of Symonds 91 (see introduction & p.4-5) says, if I am reading it correctly, that the Chern character of the representation $V$, in the sense of the formal sum of Chern classes, is

   $$ch \left( V \right) \;=\; \underset{ \mathclap{ {H_i \hookrightarrow G} \atop { W_i \in Rep(H_i) \,,\; dim(W_i) =1 } } }{\prod} \mathcal{N}_{H_i}^G ch\left(W_i\right)^{\alpha(W_i)}$$

   where

   1. the “multiplicative transfer” maps $\mathcal{N}_J^G$ are to be found in Evens 63

   2. the $\alpha(W_i)$-s are the Euler characteristics of certain CW-complexes, described on p. 3 of Symonds 91.

   So for detailed computation we next need those $\mathcal{N}$ and $\alpha$. Currently I have no feeling for how tractable these are.

   But I am thinking now that a special case of particular interest to me can now immediately be treated with this formula. But please give me a sanity check here, because my conclusion seems a little funny:

   Namely suppose we happen to take $V$ to be a virtual permutation representation (hence to be in the image of the canonical morphism $A(G) \overset{\beta}{\to} R_{\mathbb{C}} G$ from the Burnside ring).

   Since the basic permutation representations $\mathbb{C}[G/H]$ are induced from the trivial 1-dimensional $H$-representation

   $$\mathbb{C}[G/H] \; \simeq \; ind_H^G triv_{1d}$$

   it should follow that in this case the Chern characters on the right hand side of the above formula are all trivial

   $$ch(W_i) =ch(triv_{1d}) = 1 \in H^0(B G) \hookrightarrow \prod_{n} H^{2 n}(B G)$$

   ?!

   Plugging this into Symonds’ formula above would imply that all Chern classes $c_{k}$, $k \geq 1$, of virtual permutation representations just vanish.

   ?!

   Maybe I am making a mistake here, please let me know if you see where I am going astray. If this is true, it would be… interesting.

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeJan 28th 2019
   • PermaLink
   Author: Urs
   Format: MarkdownItexI have expanded a little in the entry ([here](https://ncatlab.org/nlab/show/characteristic+class+of+a+linear+representation#SplittingPrinciple)) on the above argument, making explicit that the "mutliplicative transfer" $\mathcal{N}_H^G$ is indeed multiplicative, hence sending unit Chern character to unit Chern character

   I have expanded a little in the entry (here) on the above argument, making explicit that the “mutliplicative transfer” $\mathcal{N}_H^G$ is indeed multiplicative, hence sending unit Chern character to unit Chern character

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeJan 28th 2019
   • (edited Jan 29th 2019)
   • PermaLink
   Author: Urs
   Format: MarkdownItexor rather: sends 1 to $\pm 1$. Anyway.

   or rather: sends 1 to $\pm 1$. Anyway.

9. • CommentRowNumber9.
   • CommentAuthorRichard Williamson
   • CommentTimeJan 29th 2019
   • PermaLink
   Author: Richard Williamson
   Format: MarkdownItexI have not had time to have a proper look at this yet, but it looks great! Will be interesting to see if we can compute those two things. Please continue with your ruminations, I will read when I find time :-).

   I have not had time to have a proper look at this yet, but it looks great! Will be interesting to see if we can compute those two things. Please continue with your ruminations, I will read when I find time :-).

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeJan 29th 2019
    • (edited Jan 29th 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexRight, don't worry about it until I have cleaned it all up. But if my claim here is right, then my main motivation for computing these Chern classes with the help of a computer disappears, because then we'd have general proof that the Chern classes of virtual permutation representations all vanish. At the moment this seems to me an immediate consequence of Symonds' theorem. But since this conclusion is pretty strong, I'll want to make sure that I am not overlooking something.

    Right, don’t worry about it until I have cleaned it all up.

    But if my claim here is right, then my main motivation for computing these Chern classes with the help of a computer disappears, because then we’d have general proof that the Chern classes of virtual permutation representations all vanish.

    At the moment this seems to me an immediate consequence of Symonds’ theorem. But since this conclusion is pretty strong, I’ll want to make sure that I am not overlooking something.

11. • CommentRowNumber11.
    • CommentAuthorUrs
    • CommentTimeJan 29th 2019
    • (edited Jan 29th 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexThat said, it would still be of general interest to know the Chern classes of virtual non-permutation representations. But then I don't know how much Symonds' formula can be turned into an effective algorithm that computes, since there is a fair bit of algebraic topology entering the definition of the exponents $\alpha(W_i)$ and the "multiplicative transfer" morphisms $\mathcal{N}_H^G$. This would require quite some unwinding. But let's see what we can do...

    That said, it would still be of general interest to know the Chern classes of virtual non-permutation representations. But then I don’t know how much Symonds’ formula can be turned into an effective algorithm that computes, since there is a fair bit of algebraic topology entering the definition of the exponents $\alpha(W_i)$ and the “multiplicative transfer” morphisms $\mathcal{N}_H^G$. This would require quite some unwinding. But let’s see what we can do…

12. • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeJan 29th 2019
    • (edited Jan 29th 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexI wrote in [#10](https://nforum.ncatlab.org/discussion/9527/chern-classes-of-linear-representations/?Focus=75661#Comment_75661): > But since this conclusion is pretty strong, I'll want to make sure that I am not overlooking something. Yeah, so there is a gap: We do know that a permutation representation $V$ does have a "[[Brauer induction]]" which is an expanion in induced reps of _trivial_ 1d reps, but we do not know if [Symonds' explicit Brauer induction](https://ncatlab.org/nlab/show/Brauer+induction+theorem#SymondExplicitBrauerInduction) $L(V)$ of $V$ shares this property. It might be, I suppose, that $L(V)$, being constrained by other demands, presents $V$ as a virtual combination of inductions of non-trivial 1d reps, even if for just getting any Brauer induction, trivial 1d reps would be sufficient.

    I wrote in #10:

    But since this conclusion is pretty strong, I’ll want to make sure that I am not overlooking something.

    Yeah, so there is a gap:

    We do know that a permutation representation $V$ does have a “Brauer induction” which is an expanion in induced reps of trivial 1d reps, but we do not know if Symonds’ explicit Brauer induction $L(V)$ of $V$ shares this property.

    It might be, I suppose, that $L(V)$, being constrained by other demands, presents $V$ as a virtual combination of inductions of non-trivial 1d reps, even if for just getting any Brauer induction, trivial 1d reps would be sufficient.

13. • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeJan 30th 2019
    • (edited Jan 30th 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexRight, it's actually easy to get counterexamples to that general vanishing statement. It's plain wrong, sorry for the distraction. But that's good news with regard to our project: It should (once again) be really interesting to implement an algorithm explicitly computing the Chern classes of linear reps of finite groups.

    Right, it’s actually easy to get counterexamples to that general vanishing statement. It’s plain wrong, sorry for the distraction.

    But that’s good news with regard to our project: It should (once again) be really interesting to implement an algorithm explicitly computing the Chern classes of linear reps of finite groups.

14. • CommentRowNumber14.
    • CommentAuthorRichard Williamson
    • CommentTimeJan 31st 2019
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItex> But that’s good news with regard to our project: It should (once again) be really interesting to implement an algorithm explicitly computing the Chern classes of linear reps of finite groups. Hehe, great! Sorry again, have been busy all this week with no time for nLab development or this project, but will get down to it as soon as time allows.

    But that’s good news with regard to our project: It should (once again) be really interesting to implement an algorithm explicitly computing the Chern classes of linear reps of finite groups.

    Hehe, great! Sorry again, have been busy all this week with no time for nLab development or this project, but will get down to it as soon as time allows.

15. • CommentRowNumber15.
    • CommentAuthorUrs
    • CommentTimeJan 31st 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItexThanks. No rush. Am just logging thoughts here...

    Thanks. No rush. Am just logging thoughts here…

16. • CommentRowNumber16.
    • CommentAuthorRichard Williamson
    • CommentTimeJan 31st 2019
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItexExcellent, please keep logging!

    Excellent, please keep logging!

17. • CommentRowNumber17.
    • CommentAuthorUrs
    • CommentTimeJan 31st 2019
    • (edited Jan 31st 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexHi Richard, here is a better idea of what to check next (motivated from the discussion at [[tadpole cancellation]]), and it is mathematically much simpler: Given the image of $\beta$ as a character table, we should work out inside this image the solution space to the condition $$ \chi_{V}\left( g_{\geq 1}\right) = 0 $$ hence those elements in the image whose characters vanish on all non-trivial (conjugacy classes of) group elements. For example for $G = 2 D_6$ the image of $\beta$ is spanned by these characters | $[H] = $ | $\left[\langle e\rangle\right]$ | $\left[\langle g_1\rangle\right]$ | $\left[\langle g_2\rangle\right]$ | $\left[\langle g_3\rangle\right]$ | $\left[\langle g_4\rangle\right]$ | $\left[\langle g_5\rangle\right]$ | |--|--|--|--|--|--|--| | $\chi_{V_1} =$ | $\phantom{-}1$ | $\phantom{-}1$ | $\phantom{-}1$ | $\phantom{-}1$ | $\phantom{-}1$ | $\phantom{-}1$ | | $\chi_{V_2} =$ | $\phantom{-}1$ | $\phantom{-}1$ | $\phantom{-}1$ | $-1$ | $-1$ | $\phantom{-}1$ | | $\chi_{V_3} =$ | $\phantom{-}2$ | $\phantom{-}2$ | $-1$ | $\phantom{-}0$ | $\phantom{-}0$ | $-1$ | | $\chi_{V_4} =$ | $\phantom{-}2$ | $-2$ | $\phantom{-}2$ | $\phantom{-}0$ | $\phantom{-}0$ | $-2$ | | $\chi_{V_5} =$ | $\phantom{-}4$ | $-4$ | $-2$ | $\phantom{-}0$ | $\phantom{-}0$ | $\phantom{-}2$ | and the general solution to the above condition is $$ V \;=\; n \Big( 3 V_1 + 3 V_2 + 3 V_3 + 2 V_4 + 2 V_5 \Big) \,, \phantom{AAA} n \in \mathbb{Z} $$ So the operation is simply: Remove the first column in the character table of the image of $\beta$, then solve the homogeneous linear equation determined by the transpose of the remaining matrix. The above example with $G =2D_6$ is easy to see just by eyeballing. For other examples it becomes a little tedious. For instance for $G = 2 I$ the relevant table is the one on top of p. 36 [here](https://ncatlab.org/schreiber/files/ImageofBeta190127.pdf#page=36). I just tried convincing Wolfram alpha to do this for me, by pasting that matrix into it. But, unexpectedly, Wolfram alpha keeps complaining about syntax in a way I don't presently understand.

    Hi Richard,

    here is a better idea of what to check next (motivated from the discussion at tadpole cancellation), and it is mathematically much simpler:

    Given the image of $\beta$ as a character table, we should work out inside this image the solution space to the condition

    $$\chi_{V}\left( g_{\geq 1}\right) = 0$$

    hence those elements in the image whose characters vanish on all non-trivial (conjugacy classes of) group elements.

    For example for $G = 2 D_6$ the image of $\beta$ is spanned by these characters

    $[H] =$	$\left[\langle e\rangle\right]$	$\left[\langle g_1\rangle\right]$	$\left[\langle g_2\rangle\right]$	$\left[\langle g_3\rangle\right]$	$\left[\langle g_4\rangle\right]$	$\left[\langle g_5\rangle\right]$
    $\chi_{V_1} =$	$\phantom{-}1$	$\phantom{-}1$	$\phantom{-}1$	$\phantom{-}1$	$\phantom{-}1$	$\phantom{-}1$
    $\chi_{V_2} =$	$\phantom{-}1$	$\phantom{-}1$	$\phantom{-}1$	$-1$	$-1$	$\phantom{-}1$
    $\chi_{V_3} =$	$\phantom{-}2$	$\phantom{-}2$	$-1$	$\phantom{-}0$	$\phantom{-}0$	$-1$
    $\chi_{V_4} =$	$\phantom{-}2$	$-2$	$\phantom{-}2$	$\phantom{-}0$	$\phantom{-}0$	$-2$
    $\chi_{V_5} =$	$\phantom{-}4$	$-4$	$-2$	$\phantom{-}0$	$\phantom{-}0$	$\phantom{-}2$

    and the general solution to the above condition is

    $$V \;=\; n \Big( 3 V_1 + 3 V_2 + 3 V_3 + 2 V_4 + 2 V_5 \Big) \,, \phantom{AAA} n \in \mathbb{Z}$$

    So the operation is simply: Remove the first column in the character table of the image of $\beta$, then solve the homogeneous linear equation determined by the transpose of the remaining matrix.

    The above example with $G =2D_6$ is easy to see just by eyeballing. For other examples it becomes a little tedious. For instance for $G = 2 I$ the relevant table is the one on top of p. 36 here.

    I just tried convincing Wolfram alpha to do this for me, by pasting that matrix into it. But, unexpectedly, Wolfram alpha keeps complaining about syntax in a way I don’t presently understand.

18. • CommentRowNumber18.
    • CommentAuthorDavidRoberts
    • CommentTimeJan 31st 2019
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexHmm, I tried putting the above $2D_6$ on into W|A and [got this](https://www.wolframalpha.com/input/?i=solve+%7B%7B1,1,1,1,1%7D,%7B1,1,%E2%88%921,%E2%88%921,1%7D,%7B2,%E2%88%921,0,0,%E2%88%921%7D,%7B%E2%88%922,2,0,0,%E2%88%922%7D,%7B%E2%88%924,%E2%88%922,0,0,2%7D%7D+.+%7Ba,b,c,d,e%7D+%3D+%7B%7B0%7D,%7B0%7D,%7B0%7D,%7B0%7D,%7B0%7D%7D). I may be doing it wrong, let me know.

    Hmm, I tried putting the above $2D_6$ on into W|A and got this. I may be doing it wrong, let me know.

19. • CommentRowNumber19.
    • CommentAuthorUrs
    • CommentTimeJan 31st 2019
    • (edited Jan 31st 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexYou have a transpose missing. The characters are the rows, so we want to find linear combinations of rows that vanish.

    You have a transpose missing. The characters are the rows, so we want to find linear combinations of rows that vanish.

20. • CommentRowNumber20.
    • CommentAuthorUrs
    • CommentTimeJan 31st 2019
    • (edited Jan 31st 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexBut thanks for pointing me to the right Wolfram alpha syntax. Silly me. Am on my phone now and beyond bedtime. Will get back to this tomorrow.

    But thanks for pointing me to the right Wolfram alpha syntax. Silly me. Am on my phone now and beyond bedtime. Will get back to this tomorrow.

21. • CommentRowNumber21.
    • CommentAuthorDavidRoberts
    • CommentTimeJan 31st 2019
    • (edited Jan 31st 2019)
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItex@Urs, ah, ok. I will try again. how's [this query](https://www.wolframalpha.com/input/?i=solve+Transpose%5B%7B%7B1,1,1,1,1%7D,%7B1,1,%E2%88%921,%E2%88%921,1%7D,%7B2,%E2%88%921,0,0,%E2%88%921%7D,%7B%E2%88%922,2,0,0,%E2%88%922%7D,%7B%E2%88%924,%E2%88%922,0,0,2%7D%7D%5D+.+%7Ba,b,c,d,e%7D+%3D+%7B%7B0%7D,%7B0%7D,%7B0%7D,%7B0%7D,%7B0%7D%7D) ? **Edit** no that doesn't work. But in any case, you probably want this command: NullSpace[Transpose[{{1, 1, 1, 1, 1}, {1, 1, -1, -1, 1}, {2, -1, 0, 0, -1}, {-2, 2, 0, 0, -2}, {-4, -2, 0, 0, 2}}]] which can give you [this](https://develop.open.wolframcloud.com/app/objects/f58cff69-917d-4659-8445-45355279d756) (this is straight cloud-based Mathematica, not W|A, which uses [this variant syntax](https://www.wolframalpha.com/input/?i=Nullspace+Transpose%5B%7B%7B1,1,1,1,1%7D,%7B1,1,%E2%88%921,%E2%88%921,1%7D,%7B2,%E2%88%921,0,0,%E2%88%921%7D,%7B%E2%88%922,2,0,0,%E2%88%922%7D,%7B%E2%88%924,%E2%88%922,0,0,2%7D%7D%5D)). Still debugging, though.

    @Urs, ah, ok. I will try again. how’s this query ?

    Edit no that doesn’t work. But in any case, you probably want this command:

    NullSpace[Transpose[{{1, 1, 1, 1, 1}, {1, 1, -1, -1, 1}, {2, -1, 0, 0, -1}, {-2, 2, 0, 0, -2}, {-4, -2, 0, 0, 2}}]]
    

    which can give you this (this is straight cloud-based Mathematica, not W|A, which uses this variant syntax). Still debugging, though.

22. • CommentRowNumber22.
    • CommentAuthorDavidRoberts
    • CommentTimeJan 31st 2019
    • (edited Jan 31st 2019)
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItex@Urs Your solution in #17 doesn't seem to work. In the $[\langle g_2\rangle]$ column, you get $3+3 + 3\times(-1)+2\times 2 + 2\times(-2)\neq 0$. So I think I trust Mathematica in this instance, which gives the result as the $\mathbb{Z}$-span of $V_1 + V_2 + 2V_3 + V_4+V_5$.

    @Urs Your solution in #17 doesn’t seem to work. In the $[\langle g_2\rangle]$ column, you get $3+3 + 3\times(-1)+2\times 2 + 2\times(-2)\neq 0$. So I think I trust Mathematica in this instance, which gives the result as the $\mathbb{Z}$-span of $V_1 + V_2 + 2V_3 + V_4+V_5$.

23. • CommentRowNumber23.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItexThanks for checking! Will have a look. Have you tried the case of $2I$ as in #17?

    Thanks for checking!

    Will have a look. Have you tried the case of $2I$ as in #17?

24. • CommentRowNumber24.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 1st 2019
    • (edited Feb 1st 2019)
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexNo, but [now I have](https://develop.open.wolframcloud.com/app/objects/f58cff69-917d-4659-8445-45355279d756) using the transpose of the following matrix {{1,1,1,1,1,1,1,1},{4,1,0,−1,−1,1,−1,−1},{5,−1,1,0,0,−1,0,0},{6,0,−2,1,1,0,1,1},{−12,0,0,2,2,0,−2,−2},{−8,2,0,−2,−2,−2,2,2},{−8,−4,0,−2,−2,4,2,2}} and if I've got that right, the coefficient vector is $(1,4,5,3,3,2,1)$

    No, but now I have using the transpose of the following matrix

    {{1,1,1,1,1,1,1,1},{4,1,0,−1,−1,1,−1,−1},{5,−1,1,0,0,−1,0,0},{6,0,−2,1,1,0,1,1},{−12,0,0,2,2,0,−2,−2},{−8,2,0,−2,−2,−2,2,2},{−8,−4,0,−2,−2,4,2,2}}
    

    and if I’ve got that right, the coefficient vector is $(1,4,5,3,3,2,1)$

25. • CommentRowNumber25.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • (edited Feb 1st 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexThanks! Am still kind of offline here, but that's interesting thar for $2I$ there is still a $\mathbb{Z}$ worth of solutions, since this case has again one more constraint than variables. Remains then only to check the case of $2O$...

    Thanks!

    Am still kind of offline here, but that’s interesting thar for $2I$ there is still a $\mathbb{Z}$ worth of solutions, since this case has again one more constraint than variables.

    Remains then only to check the case of $2O$…

26. • CommentRowNumber26.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 1st 2019
    • (edited Feb 1st 2019)
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexYes, was thinking of doing it :-) again, one-dimensional (1,1,2,3,3,2,1), calculation [here](https://develop.open.wolframcloud.com/app/objects/f58cff69-917d-4659-8445-45355279d756), using NullSpace[Transpose[{{1,1,1,1,1,1,1},{1,1,1,−1,1,−1,−1},{2,−1,2,0,−1,0,0},{3,0,−1,1,0,−1,−1},{3,0,−1,−1,0,1,1},{−8,2,0,0,−2,0,0},{−8,−4,0,0,4,0,0}}]]

    Yes, was thinking of doing it :-) again, one-dimensional (1,1,2,3,3,2,1), calculation here, using

    NullSpace[Transpose[{{1,1,1,1,1,1,1},{1,1,1,−1,1,−1,−1},{2,−1,2,0,−1,0,0},{3,0,−1,1,0,−1,−1},{3,0,−1,−1,0,1,1},{−8,2,0,0,−2,0,0},{−8,−4,0,0,4,0,0}}]]
    

27. • CommentRowNumber27.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 1st 2019
    • (edited Feb 1st 2019)
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItex[deleted]

    [deleted]

28. • CommentRowNumber28.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • (edited Feb 1st 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexThanks once more! (Hope to finally get going now. It's the analog of Sunday here...)

    Thanks once more! (Hope to finally get going now. It’s the analog of Sunday here…)

29. • CommentRowNumber29.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • (edited Feb 1st 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexOkay, I have fixed [here](https://ncatlab.org/nlab/show/RR-field+tadpole+cancellation#At2D6Singularity) the example of $2 D_6$ as you pointed out in [#22](https://nforum.ncatlab.org/discussion/9527/chern-classes-of-linear-representations/?Focus=75739#Comment_75739). That makes the mass in this case be $M_{2 D_6} =10$. I also found an arithmetic mistake in the computation of the mass in the case of $2 D_8$ [here](https://ncatlab.org/nlab/show/RR-field+tadpole+cancellation#At2D8Singularity). The correct value seems to be $M_{2 D_8} =14$. Now I wanted to see if there is a pattern in how the mass grows, so I added also the example of $2 D_{10}$, [here](https://ncatlab.org/nlab/show/RR-field+tadpole+cancellation#At2D10Singularity) (again just by eyeballing, which, as we have seen, is error prone, simplistic as the problem may be) So now I seem to have this (non-)pattern: $$ \begin{aligned} M_{2 D_4} & = 8 \\ M_{2 D_6} & = 10 \\ M_{2 D_8} & = 14 \\ M_{2 D_{10}} & = 20 \end{aligned} $$ Hm, so maybe $M_{2 D_{{2(k+1) + 2}}} = M_{2 D_{{2 k + 2}}} + 2k$, generally.

    Okay, I have fixed here the example of $2 D_6$ as you pointed out in #22.

    That makes the mass in this case be $M_{2 D_6} =10$.

    I also found an arithmetic mistake in the computation of the mass in the case of $2 D_8$ here. The correct value seems to be $M_{2 D_8} =14$.

    Now I wanted to see if there is a pattern in how the mass grows, so I added also the example of $2 D_{10}$, here (again just by eyeballing, which, as we have seen, is error prone, simplistic as the problem may be)

    So now I seem to have this (non-)pattern:

    $$\begin{aligned} M_{2 D_4} & = 8 \\ M_{2 D_6} & = 10 \\ M_{2 D_8} & = 14 \\ M_{2 D_{10}} & = 20 \end{aligned}$$

    Hm, so maybe $M_{2 D_{{2(k+1) + 2}}} = M_{2 D_{{2 k + 2}}} + 2k$, generally.

30. • CommentRowNumber30.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItexHave computed the case of $2 D_{12}$ now, [here](https://ncatlab.org/nlab/show/RR-field+tadpole+cancellation#At2D12Singularity). I get $$ M_{2 D_{12}} = 24 \,. $$ If that's correct, it breaks the pattern that i guessed above (not that there is any reason there should be such a pattern, it just seemed suggestive).

    Have computed the case of $2 D_{12}$ now, here. I get

    $$M_{2 D_{12}} = 24 \,.$$

    If that’s correct, it breaks the pattern that i guessed above (not that there is any reason there should be such a pattern, it just seemed suggestive).

31. • CommentRowNumber31.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • (edited Feb 1st 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexhave computed the case $2 D_{14}$ now, [here](https://ncatlab.org/nlab/show/RR-field+tadpole+cancellation#At2D14Singularity). I get $$ M_{2 D_{14}} = 28 $$

    have computed the case $2 D_{14}$ now, here. I get

    $$M_{2 D_{14}} = 28$$
32. • CommentRowNumber32.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItexadded the case of $2 D_{16}$, [here](https://ncatlab.org/nlab/show/RR-field+tadpole+cancellation#At2D16Singularity) That made it clear what the mass pattern wants to be, and so I went back, looked for and found further arithmetic errors. Now it's clear: | [[singularity]] | [[mass]] | |----|-----| | $2 D_4$ | $8$ | | $2 D_6$ | $12$ | | $2 D_{8}$ | $16$ | | $2 D_{10}$ | $20$ | | $2 D_{12}$ | $24$ | | $2 D_{14}$ | $28$ | | $2 D_{16}$ | $32$ |

    added the case of $2 D_{16}$, here

    That made it clear what the mass pattern wants to be, and so I went back, looked for and found further arithmetic errors. Now it’s clear:

    singularity	mass
    $2 D_4$	$8$
    $2 D_6$	$12$
    $2 D_{8}$	$16$
    $2 D_{10}$	$20$
    $2 D_{12}$	$24$
    $2 D_{14}$	$28$
    $2 D_{16}$	$32$

33. • CommentRowNumber33.
    • CommentAuthorRichard Williamson
    • CommentTimeFeb 1st 2019
    • PermaLink
    Author: Richard Williamson
    Format: MarkdownItexIs it possible to say a word for a physics layman like myself as to what such a pattern could signify?

    Is it possible to say a word for a physics layman like myself as to what such a pattern could signify?

34. • CommentRowNumber34.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 1st 2019
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexHmm, curious! So the other masses, if I have mental arithmetic right, are: * 2T - 24 * 2O - 48 * 2I - 120 Hence _also_ the orders of these groups.

    Hmm, curious!

    So the other masses, if I have mental arithmetic right, are:

    • 2T - 24
    • 2O - 48
    • 2I - 120

    Hence also the orders of these groups.

35. • CommentRowNumber35.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • (edited Feb 1st 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexRichard asks in #33: > Is it possible to say a word for a physics layman like myself as to what such a pattern could signify? Yes, so this is saying that the unit mass of the [[RR-field tadpole cancellation|RR-chargeless]] [[fractional D-branes]] which are stuck at a $2 D_{2k}$-orbifold singularity increases linearly with $k$ as $M_{2 D_{2k}} = 4k$. These orbifold singularities we may think of as being higher dimensional extremal black hole singularities viewed from larger than Planck scale distance (by the discussion at [[near-horizon geometry]]) and so this says that the "deeper" the singularity, the more massive the corresponding stuff inside there is. Which seems to make good sense.

    Richard asks in #33:

    Is it possible to say a word for a physics layman like myself as to what such a pattern could signify?

    Yes, so this is saying that the unit mass of the RR-chargeless fractional D-branes which are stuck at a $2 D_{2k}$-orbifold singularity increases linearly with $k$ as $M_{2 D_{2k}} = 4k$.

    These orbifold singularities we may think of as being higher dimensional extremal black hole singularities viewed from larger than Planck scale distance (by the discussion at near-horizon geometry) and so this says that the “deeper” the singularity, the more massive the corresponding stuff inside there is. Which seems to make good sense.

36. • CommentRowNumber36.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItexDavid says in #34: > Hence _also_ the orders of these groups. Thanks. Yes, that's really interesting now! I wonder if the representation theorists would recognize what is going on here as something known. Will type out the $2O$ and $2I$ case now, just so that we have it all in front of us...

    David says in #34:

    Hence also the orders of these groups.

    Thanks. Yes, that’s really interesting now!

    I wonder if the representation theorists would recognize what is going on here as something known.

    Will type out the $2O$ and $2I$ case now, just so that we have it all in front of us…

37. • CommentRowNumber37.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItexAh, it's even better: There is a normalization factor $1/{\vert G \vert }$ which converts from character values to fractional D-brane charge ([here](https://ncatlab.org/nlab/show/fractional+D-brane#eq:RRChargeOfFractionalDBraneInGTwistedSector)). This means that we find in all cases that the $\mathbb{Z}$-sublattice of RR-charge-free fractional branes is always spanned by a brane of unit mass.

    Ah, it’s even better:

    There is a normalization factor $1/{\vert G \vert }$ which converts from character values to fractional D-brane charge (here).

    This means that we find in all cases that the $\mathbb{Z}$-sublattice of RR-charge-free fractional branes is always spanned by a brane of unit mass.

38. • CommentRowNumber38.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 1st 2019
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexVery nice!

    Very nice!

39. • CommentRowNumber39.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • (edited Feb 1st 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexI should repeat at this point the warning that I mentioned in the other thread [here](https://nforum.ncatlab.org/discussion/8433/rrfield-tadpole-cancellation/?Focus=75713#Comment_75713): The string theory literature tends to focus on the A-type singularities (just because it's easier), whereas here we are focusing on the complement, the D- and E-type singularities (because for these we know that the virtual permutation reps span the non-irrational charge sublattice). So that $1/{\vert G\vert}$-conversion formula, as well as the tadpole cancellation condition $\chi_{V}(g_{\geq 1}) = 0$ itself, is all derived in the literature for A-type singularities. While, by their pure representation-theoretic nature, these formulas have an evident and immediate generalization to other types of singularities, such as the ones we are considering here, it would be good to find some reference that confirms that in this more general case these formulas indeed still follow from the corresponding string-scattering computations. I haven't seen such a reference yet. I have emailed some people about this, but not heard back yet.

    I should repeat at this point the warning that I mentioned in the other thread here:

    The string theory literature tends to focus on the A-type singularities (just because it’s easier), whereas here we are focusing on the complement, the D- and E-type singularities (because for these we know that the virtual permutation reps span the non-irrational charge sublattice). So that $1/{\vert G\vert}$-conversion formula, as well as the tadpole cancellation condition $\chi_{V}(g_{\geq 1}) = 0$ itself, is all derived in the literature for A-type singularities. While, by their pure representation-theoretic nature, these formulas have an evident and immediate generalization to other types of singularities, such as the ones we are considering here, it would be good to find some reference that confirms that in this more general case these formulas indeed still follow from the corresponding string-scattering computations. I haven’t seen such a reference yet. I have emailed some people about this, but not heard back yet.

40. • CommentRowNumber40.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItexNow I have spelled out the last case, of $2I$, [here](https://ncatlab.org/nlab/show/RR-field+tadpole+cancellation#At2ISingularity). David, could you be so kind to look over this briefly? Because I am taking this on faith from your comment #24, find it too cumbersome to check this mentally...

    Now I have spelled out the last case, of $2I$, here.

    David, could you be so kind to look over this briefly? Because I am taking this on faith from your comment #24, find it too cumbersome to check this mentally…

41. • CommentRowNumber41.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 1st 2019
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexYes, the $2I$ case is fine. I double checked that I'd used the correct matrix in the Mathematica calculation, and that your mass calculation is correct following on from that result.

    Yes, the $2I$ case is fine. I double checked that I’d used the correct matrix in the Mathematica calculation, and that your mass calculation is correct following on from that result.

42. • CommentRowNumber42.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • (edited Feb 1st 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexThanks, David!! Here is an observation: The result that we find unit normalized mass everywhere, implies something interesting in view of the following general formula for plain (non-virtual) representations: $$ dim\left( V^G \right) \;=\; \frac{1}{{\vert G\vert}} \underset{g \in G}{\sum} \chi_V(g) $$ (of course elementary, basic and standard, but e.g. (2.1) [here](https://www.uni-math.gwdg.de/tammo/rep.pdf)) Now applying this to our virtual reps $V_+ - V_-$ with their defining property and that unit mass result implies that $$ dim\left( (V_+)^G \right) - dim\left( (V_-)^G \right) \;=\; 1 $$ Not sure if this is of any relevance. But it seems curious.

    Thanks, David!!

    Here is an observation:

    The result that we find unit normalized mass everywhere, implies something interesting in view of the following general formula for plain (non-virtual) representations:

    $$dim\left( V^G \right) \;=\; \frac{1}{{\vert G\vert}} \underset{g \in G}{\sum} \chi_V(g)$$

    (of course elementary, basic and standard, but e.g. (2.1) here)

    Now applying this to our virtual reps $V_+ - V_-$ with their defining property and that unit mass result implies that

    $$dim\left( (V_+)^G \right) - dim\left( (V_-)^G \right) \;=\; 1$$

    Not sure if this is of any relevance. But it seems curious.

43. • CommentRowNumber43.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 1st 2019
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexSeems like an index-theoretic result to me.

    Seems like an index-theoretic result to me.

44. • CommentRowNumber44.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • (edited Feb 1st 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexTrue, but I would say that's just because it is a statement about a virtual rep. The tadpole condition we have been discussing also looks like index theory in this way. But what I found striking here is that the tadpole condition (which is something about all group elements separately) together with the unit mass result gives a global statement about the fixed point locus of the whole group. This might be pointing to some nice abstract condition in equivariant homotopy theory. Not sure.

    True, but I would say that’s just because it is a statement about a virtual rep. The tadpole condition we have been discussing also looks like index theory in this way.

    But what I found striking here is that the tadpole condition (which is something about all group elements separately) together with the unit mass result gives a global statement about the fixed point locus of the whole group. This might be pointing to some nice abstract condition in equivariant homotopy theory. Not sure.

45. • CommentRowNumber45.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItexWait. I think all we did here was express the regular representation in an unfamiliar basis...

    Wait. I think all we did here was express the regular representation in an unfamiliar basis…

46. • CommentRowNumber46.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 1st 2019
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexNot a terrible thing, a well-chosen basis can make all the difference.

    Not a terrible thing, a well-chosen basis can make all the difference.

47. • CommentRowNumber47.
    • CommentAuthorUrs
    • CommentTimeFeb 2nd 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItexRight, it's not terrible, but there is something strange. I'll upload the integral character table for some cyclic groups, and then let's check if the kernel of the columns away from the neutral element is still just 1-dimensional. If it is, there is something odd with the literature, if it isn't, that would be interesting

    Right, it’s not terrible, but there is something strange. I’ll upload the integral character table for some cyclic groups, and then let’s check if the kernel of the columns away from the neutral element is still just 1-dimensional. If it is, there is something odd with the literature, if it isn’t, that would be interesting

48. • CommentRowNumber48.
    • CommentAuthorUrs
    • CommentTimeFeb 2nd 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItexMeanwhile, could you just confirm that the solution space in all examples we discussed here is really just 1-dimensional?

    Meanwhile, could you just confirm that the solution space in all examples we discussed here is really just 1-dimensional?

49. • CommentRowNumber49.
    • CommentAuthorUrs
    • CommentTimeFeb 2nd 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItexSo we are really asking: For which groups may we forget about the first column of the character table, the dimensions, and still have that the character map is injective? (Luckily I have no trouble showing off my ignorance in basic representation theory :-)

    So we are really asking: For which groups may we forget about the first column of the character table, the dimensions, and still have that the character map is injective?

    (Luckily I have no trouble showing off my ignorance in basic representation theory :-)

50. • CommentRowNumber50.
    • CommentAuthorUrs
    • CommentTimeFeb 2nd 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItex(Never mind the confirmation, I have now linked all examples [here](https://ncatlab.org/nlab/show/RR-field+tadpole+cancellation#examples) to the respective Wolfram alpha computation output.)

    (Never mind the confirmation, I have now linked all examples here to the respective Wolfram alpha computation output.)

51. • CommentRowNumber51.
    • CommentAuthorUrs
    • CommentTimeFeb 2nd 2019
    • (edited Feb 2nd 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexWhat do I need to type to convince Wolfram alpha to consider primitive roots of unity inside a matrix? (e.g. [this](https://www.wolframalpha.com/input/?i=Nullspace%5B%7B%7BExp%5B2+Pi+I+%2F3%5D%7D%7D%5D) isn't recognized, why not?)

    What do I need to type to convince Wolfram alpha to consider primitive roots of unity inside a matrix?

    (e.g. this isn’t recognized, why not?)

52. • CommentRowNumber52.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 3rd 2019
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexNo idea! I tried [this](https://www.wolframalpha.com/input/?i=Nullspace%5B%7B%7BExp%5B2*Pi*I%2F3%5D,1%7D,%7B2*Exp%5B2*Pi*I%2F3%5D,2%7D%7D%5D) after some experimentation and it worked.

    No idea! I tried this after some experimentation and it worked.

53. • CommentRowNumber53.
    • CommentAuthorUrs
    • CommentTimeFeb 3rd 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItexThanks! Meanwhile, I found the general argument that multiples of the regular rep are the only solution [here](https://nforum.ncatlab.org/discussion/8433/rrfield-tadpole-cancellation/?Focus=75805#Comment_75805) (completely elementary, of course). Now I am thinking about how to contact these authors to alert them of their study of the empty set. Or of the singleton, rather.

    Thanks!

    Meanwhile, I found the general argument that multiples of the regular rep are the only solution here (completely elementary, of course). Now I am thinking about how to contact these authors to alert them of their study of the empty set. Or of the singleton, rather.

54. • CommentRowNumber54.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 3rd 2019
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexSo those authors only consider the abelian (i.e $A$-type subgroups of $SU(2)$) case? Does your result make extension of their analysis to the $D$ and $E$ subgroups more or less plausible?

    So those authors only consider the abelian (i.e $A$-type subgroups of $SU(2)$) case? Does your result make extension of their analysis to the $D$ and $E$ subgroups more or less plausible?

55. • CommentRowNumber55.
    • CommentAuthorUrs
    • CommentTimeFeb 3rd 2019
    • PermaLink
    Author: Urs
    Format: MarkdownItexLooking at section 2.2. of Honecker 02 [arXiv:hep-th/0201037](https://arxiv.org/abs/hep-th/0201037) (which is what section 4 of Marchesano 04 [arXiv:hep-th/0307252](https://arxiv.org/abs/hep-th/0307252) is largely based on) it seems like there is really no assumption on the nature of the orbifold group necessary. It's just taken to be cyclic at the end of the computation, because that happens to be relevant in the model that the rest of the article is concerned with. To clarify, I have sent another email to the authors, turns out my first attempt bounced.

    Looking at section 2.2. of Honecker 02 arXiv:hep-th/0201037 (which is what section 4 of Marchesano 04 arXiv:hep-th/0307252 is largely based on) it seems like there is really no assumption on the nature of the orbifold group necessary. It’s just taken to be cyclic at the end of the computation, because that happens to be relevant in the model that the rest of the article is concerned with.

    To clarify, I have sent another email to the authors, turns out my first attempt bounced.

56. • CommentRowNumber56.
    • CommentAuthorUrs
    • CommentTimeFeb 3rd 2019
    • (edited Feb 3rd 2019)
    • PermaLink
    Author: Urs
    Format: MarkdownItexBut I take back my statement about studying the empty set/singleton set. It's true that only multiples of the regular rep solve the representation-theoretic factor of those tadpole cancellation conditions, but from (4.9), (4.12), (4.14) in Marchesano's text, of course the point now is to tensor the equivariant K-theory with another (co-)homology group and then ask something like that the elements in this tensor product end up componentwise in the tensor product with just the multiples of regular reps. Or something like this. Need to find a good abstract way of capturing these conditions...

    But I take back my statement about studying the empty set/singleton set. It’s true that only multiples of the regular rep solve the representation-theoretic factor of those tadpole cancellation conditions, but from (4.9), (4.12), (4.14) in Marchesano’s text, of course the point now is to tensor the equivariant K-theory with another (co-)homology group and then ask something like that the elements in this tensor product end up componentwise in the tensor product with just the multiples of regular reps. Or something like this. Need to find a good abstract way of capturing these conditions…