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1. • CommentRowNumber1.
   • CommentAuthorAndrew Stacey
   • CommentTimeApr 29th 2013
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexListening to a talk recently on trace-class operators on locally convex topological vector spaces (so, not assumed to be Hilbert spaces - though Banach would do for what I'm going to describe), I ended up wondering whether or not the following was an example of a standard categorical construction. The idea is to have something like a category except that there are two types of morphism between the objects. Naïvely, the definition might go like this: 1. A class of objects, say $O$ 1. For each pair of objects, say $a,b \in O$, two sets $M(a,b)$ and $N(a,b)$ such that: 1. The data $(O,M(-,-))$ forms a category, 2. The data $(O,N(-,-))$ satisfies the composition law but need not have identities, 3. $(O,M(-,-))$ acts on $(O,N(-,-))$ in that there is a product $M(a,b) \times N(b,c) \to N(a,c)$ satisfying all the obvious coherences. The motivating example being where the objects are locally convex topological vector spaces, $M(a,b)$ are continuous linear maps, and $N(a,b)$ are trace-class maps. Note that the identity is rarely trace-class so that excludes $(O,N(-,-))$ from being a category, and that the natural map from trace-class to all is not necessarily injective so this isn't a category with distinguished morphisms. One could also formulate cobordisms in this way, with diffeomorphisms as the $M(-,-)$ morphisms, since one has to do a bit of fiddling to make the identity map a cobordism.

   Listening to a talk recently on trace-class operators on locally convex topological vector spaces (so, not assumed to be Hilbert spaces - though Banach would do for what I’m going to describe), I ended up wondering whether or not the following was an example of a standard categorical construction.

   The idea is to have something like a category except that there are two types of morphism between the objects. Naïvely, the definition might go like this:

   1. A class of objects, say $O$
   2. For each pair of objects, say $a,b \in O$, two sets $M(a,b)$ and $N(a,b)$

   such that:

   1. The data $(O,M(-,-))$ forms a category,
   2. The data $(O,N(-,-))$ satisfies the composition law but need not have identities,
   3. $(O,M(-,-))$ acts on $(O,N(-,-))$ in that there is a product $M(a,b) \times N(b,c) \to N(a,c)$ satisfying all the obvious coherences.

   The motivating example being where the objects are locally convex topological vector spaces, $M(a,b)$ are continuous linear maps, and $N(a,b)$ are trace-class maps. Note that the identity is rarely trace-class so that excludes $(O,N(-,-))$ from being a category, and that the natural map from trace-class to all is not necessarily injective so this isn’t a category with distinguished morphisms.

   One could also formulate cobordisms in this way, with diffeomorphisms as the $M(-,-)$ morphisms, since one has to do a bit of fiddling to make the identity map a cobordism.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeApr 29th 2013
   • PermaLink
   Author: Urs
   Format: MarkdownItexAn [[ideal]] in the monoid-oid of morphisms, hence a [[sieve]]. Isn't that what you describe?

   An ideal in the monoid-oid of morphisms, hence a sieve. Isn’t that what you describe?

3. • CommentRowNumber3.
   • CommentAuthorAndrew Stacey
   • CommentTimeApr 29th 2013
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexClose, but fails on two counts which are exhibited by the motivating example of trace-class operators: 1. Trace-class operators are not always a subset of all operators, so the "category" $(O,N(-,-))$ is not a subcategory of $(O,M(-,-))$. 1. The identity is rarely a trace-class operators, so the "category" $(O,N(-,-))$ lacks identities.

   Close, but fails on two counts which are exhibited by the motivating example of trace-class operators:

   1. Trace-class operators are not always a subset of all operators, so the “category” $(O,N(-,-))$ is not a subcategory of $(O,M(-,-))$.
   2. The identity is rarely a trace-class operators, so the “category” $(O,N(-,-))$ lacks identities.
4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeApr 29th 2013
   • (edited Apr 29th 2013)
   • PermaLink
   Author: Urs
   Format: MarkdownItexRe 1: You consider trace-class operators which are not linear and/or not continuous? Re 2: if $N$ is or were a sieve inside $M$, then it need not have identities.

   Re 1: You consider trace-class operators which are not linear and/or not continuous?

   Re 2: if $N$ is or were a sieve inside $M$, then it need not have identities.

5. • CommentRowNumber5.
   • CommentAuthorAndrew Stacey
   • CommentTimeApr 29th 2013
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexRe 1: I should be more explicit in what I mean by "trace-class operator". I mean an element in the trace-closure of the tensor product of $a' \otimes b$. I guess I was a bit lax here, sorry. Re 2: Okay, that's not what the page [[sieve]] says: > A sieve in a category C is a full subcategory closed under precomposition with morphisms in C. The "full subcategory" seems pretty clear to me that it contains identities.

   Re 1: I should be more explicit in what I mean by “trace-class operator”. I mean an element in the trace-closure of the tensor product of $a' \otimes b$. I guess I was a bit lax here, sorry.

   Re 2: Okay, that’s not what the page sieve says:

   A sieve in a category C is a full subcategory closed under precomposition with morphisms in C.

   The “full subcategory” seems pretty clear to me that it contains identities.

6. • CommentRowNumber6.
   • CommentAuthorzskoda
   • CommentTimeApr 29th 2013
   • (edited Apr 29th 2013)
   • PermaLink
   Author: zskoda
   Format: MarkdownItex5 -- about sieve -- this definition is strange...ideal in a monoid, an $\Omega$-group, and in particular in a unital ring is not having a multiplicative identity (otherwise it is the whole ring), hence it is not a unital subring. Probably, the page [[sieve]] has it wrong...

   5 – about sieve – this definition is strange…ideal in a monoid, an $\Omega$-group, and in particular in a unital ring is not having a multiplicative identity (otherwise it is the whole ring), hence it is not a unital subring. Probably, the page sieve has it wrong…

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeApr 29th 2013
   • (edited Apr 29th 2013)
   • PermaLink
   Author: Urs
   Format: MarkdownItex@Andrew, Zoran: Go one line further in _[[sieve]]_. (I guess that Idea-section could do with a better lead-in, but I won't look into that right now.)

   @Andrew, Zoran: Go one line further in sieve.

   (I guess that Idea-section could do with a better lead-in, but I won’t look into that right now.)

8. • CommentRowNumber8.
   • CommentAuthorzskoda
   • CommentTimeApr 29th 2013
   • (edited Apr 29th 2013)
   • PermaLink
   Author: zskoda
   Format: MarkdownItexSo, is the sieve in the sense of that definition the same as an ideal in a category or not ? (I understand that the slice category version works well fitting with the terminology for [[Grothendieck topologies]], besides I read that exposition in Lurie some time ago). I am having in mind predominantly the context of enriched categories where the case of an enriched category with one object should be a special case, hence a ring as an one object version of an additive category.

   So, is the sieve in the sense of that definition the same as an ideal in a category or not ? (I understand that the slice category version works well fitting with the terminology for Grothendieck topologies, besides I read that exposition in Lurie some time ago). I am having in mind predominantly the context of enriched categories where the case of an enriched category with one object should be a special case, hence a ring as an one object version of an additive category.

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeApr 29th 2013
   • PermaLink
   Author: Urs
   Format: MarkdownItexA sieve over an object is a right ideal: a collection of maps into that object which is closed under precompposition.

   A sieve over an object is a right ideal: a collection of maps into that object which is closed under precompposition.

10. • CommentRowNumber10.
    • CommentAuthorzskoda
    • CommentTimeApr 29th 2013
    • PermaLink
    Author: zskoda
    Format: MarkdownItexI understand the "right" modifier, but still, for additive category with one object I do not see how do you exclude the identity (if you include it, you get everything, hence one has only one such right ideal for one object additive category...).

    I understand the “right” modifier, but still, for additive category with one object I do not see how do you exclude the identity (if you include it, you get everything, hence one has only one such right ideal for one object additive category…).

11. • CommentRowNumber11.
    • CommentAuthorUrs
    • CommentTimeApr 29th 2013
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    Author: Urs
    Format: MarkdownItexIt almost seems as if we are discussing what an ideal is and what a sieve is. Is that right? :-) Zoran, are you asking the following: given a ring $R$ and an ideal $N \hookrightarrow R$, why does that define a sieve over the unique object of $\mathbf{B}R$?

    It almost seems as if we are discussing what an ideal is and what a sieve is. Is that right? :-)

    Zoran, are you asking the following:

    given a ring $R$ and an ideal $N \hookrightarrow R$, why does that define a sieve over the unique object of $\mathbf{B}R$?

12. • CommentRowNumber12.
    • CommentAuthorzskoda
    • CommentTimeApr 29th 2013
    • (edited Apr 29th 2013)
    • PermaLink
    Author: zskoda
    Format: MarkdownItexI am asking how come given the _proper_ ideal $N\hookrightarrow R$, which clearly does not have an identity element from $R$, what then with the identity morphism for the unique object in $\mathbf{B}R$, which should be in the corresponding sieve (the same elements as morphisms of right multiplication), if sieve is a subcategory, as the entry says? What am I misreading ?

    I am asking how come given the proper ideal $N\hookrightarrow R$, which clearly does not have an identity element from $R$, what then with the identity morphism for the unique object in $\mathbf{B}R$, which should be in the corresponding sieve (the same elements as morphisms of right multiplication), if sieve is a subcategory, as the entry says? What am I misreading ?

13. • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeApr 29th 2013
    • PermaLink
    Author: Urs
    Format: MarkdownItexZoran, please read beyond the first line of the entry. It says (and I have now amplified this more in the new version) that a sieve on an object is a full subcategory closed under precomposition in the slice over it.

    Zoran,

    please read beyond the first line of the entry. It says (and I have now amplified this more in the new version) that a sieve on an object is a full subcategory closed under precomposition in the slice over it.

14. • CommentRowNumber14.
    • CommentAuthorzskoda
    • CommentTimeApr 29th 2013
    • (edited Apr 29th 2013)
    • PermaLink
    Author: zskoda
    Format: MarkdownItexI see now -- my confusion, prompted by 2 was that "ideal" suggestion and in categories I think of ideals quite differently (as a congruence on a category, not in any of the senses covered in $n$Lab, it seems), than of the sieves (strangely, the entry [[ideal]] has nothing for the case of additive categories).

    I see now – my confusion, prompted by 2 was that “ideal” suggestion and in categories I think of ideals quite differently (as a congruence on a category, not in any of the senses covered in $n$Lab, it seems), than of the sieves (strangely, the entry ideal has nothing for the case of additive categories).

15. • CommentRowNumber15.
    • CommentAuthorUrs
    • CommentTimeApr 29th 2013
    • PermaLink
    Author: Urs
    Format: MarkdownItexToby has written these entries in quite some generality. The entry closer to the standard notion is _[[ideal in a monoid]]_.

    Toby has written these entries in quite some generality. The entry closer to the standard notion is ideal in a monoid.

16. • CommentRowNumber16.
    • CommentAuthorzskoda
    • CommentTimeApr 29th 2013
    • PermaLink
    Author: zskoda
    Format: MarkdownItex> in quite some generality I meant the obvious: for categories as oidification of monoids.

    in quite some generality

    I meant the obvious: for categories as oidification of monoids.

17. • CommentRowNumber17.
    • CommentAuthorUrs
    • CommentTimeApr 29th 2013
    • PermaLink
    Author: Urs
    Format: MarkdownItex@Zoran, not sure what you mean that you meant. Sorry. Can you just say again which of the various statements that we talked about you are making which statement about? Sorry. @Andrew: to answer your question in the case that $N$ is not a subcategory of $M$: then what you have is still an _[[action of a category]]_ of $M$ an $N$.

    @Zoran, not sure what you mean that you meant. Sorry. Can you just say again which of the various statements that we talked about you are making which statement about? Sorry.

    @Andrew: to answer your question in the case that $N$ is not a subcategory of $M$: then what you have is still an action of a category of $M$ an $N$.

18. • CommentRowNumber18.
    • CommentAuthorKarol Szumiło
    • CommentTimeApr 30th 2013
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    Author: Karol Szumiło
    Format: MarkdownItexThose look like categories enriched over a particular monoidal category, namely, the category of pairs of sets $(M, N)$ with the product $(M_0, N_0) \otimes (M_1, N_1) = (M_0 \times M_1, M_0 \times N_1)$ and the unit $(*, \varnothing)$. This is an instance of a Day convolution product, but only over a [[promonoidal category]], not an honest monoidal category. Moreover, this monoidal category is not symmetric and the theory of enriched categories over it may be a little hairy so I'm not sure how useful this observation is.

    Those look like categories enriched over a particular monoidal category, namely, the category of pairs of sets $(M, N)$ with the product $(M_0, N_0) \otimes (M_1, N_1) = (M_0 \times M_1, M_0 \times N_1)$ and the unit $(*, \varnothing)$. This is an instance of a Day convolution product, but only over a promonoidal category, not an honest monoidal category. Moreover, this monoidal category is not symmetric and the theory of enriched categories over it may be a little hairy so I’m not sure how useful this observation is.

19. • CommentRowNumber19.
    • CommentAuthorTobyBartels
    • CommentTimeApr 30th 2013
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItexSince $N \nsubseteq M$, we have more of a module than an ideal?

    Since $N \nsubseteq M$, we have more of a module than an ideal?

20. • CommentRowNumber20.
    • CommentAuthorAndrew Stacey
    • CommentTimeApr 30th 2013
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexKarol, Thanks! That's not quite it, but it's given me the clue as to what it is. I had an inkling that there would be some sort of enriched description of this but (as I'm no category theorist) had no idea how to formulate it. I've been wanting to say "enriched over "Groups $\times$ $G$-sets" except that that only describes the automorphisms and not the morphisms and I didn't know how to unroll that to a proper description. Seeing what you wrote showed me how to do that. The category that it is enriched over is the category of pairs of sets, but the product is actually: $$ (M_0,N_0) \otimes (M_1, N_1) = (M_0 \times M_1, M_0 \times N_1 \amalg N_0 \times N_1 \amalg N_0 \times M_1) $$ I guess one could simply make this an ordinary category by folding the pairs into a coproduct, but I'd hope that there was some gain to regarding it in this enriched way. This product Is symmetric so that should make life easier - or does your dire warning still hold?

    Karol, Thanks! That’s not quite it, but it’s given me the clue as to what it is. I had an inkling that there would be some sort of enriched description of this but (as I’m no category theorist) had no idea how to formulate it. I’ve been wanting to say “enriched over “Groups $\times$ $G$-sets” except that that only describes the automorphisms and not the morphisms and I didn’t know how to unroll that to a proper description. Seeing what you wrote showed me how to do that.

    The category that it is enriched over is the category of pairs of sets, but the product is actually:

    $$(M_0,N_0) \otimes (M_1, N_1) = (M_0 \times M_1, M_0 \times N_1 \amalg N_0 \times N_1 \amalg N_0 \times M_1)$$

    I guess one could simply make this an ordinary category by folding the pairs into a coproduct, but I’d hope that there was some gain to regarding it in this enriched way.

    This product Is symmetric so that should make life easier - or does your dire warning still hold?

21. • CommentRowNumber21.
    • CommentAuthorKarol Szumiło
    • CommentTimeApr 30th 2013
    • PermaLink
    Author: Karol Szumiło
    Format: MarkdownItexYou are right, I've failed to take into account that N-morphisms also have composition. And from your third summand I gather that you want M-morphisms to act on N-morphisms from both sides. This indeed makes the monoidal structure symmetric. In fact, now it has a very nice description. Let $E = \{ 1, e \}$ be a monoid with $1$ a unit and $e$ an idempotent. Consider it as a monoidal category with elements of $E$ as objects, no non-trivial morphisms and the monoidal product induced by the multiplication of $E$. Then your monoidal product is the Day convolution of functors $E \to \mathrm{Set}$. Since $E$ is commutative this makes $\mathrm{Set}^E$ into a closed symmetric monoidal category. It is also complete and cocomplete (even locally presentable) so all the standard theory of enriched categories is available. I think it is more natural (and more categorical) to think of objects in this category as pairs of sets. (Just as it is more natural to think of graded modules and rings as sequences of abelian groups as opposed to abelian groups equipped with a decomposition into a direct sum of a sequence of subgroups.)

    You are right, I’ve failed to take into account that N-morphisms also have composition. And from your third summand I gather that you want M-morphisms to act on N-morphisms from both sides. This indeed makes the monoidal structure symmetric. In fact, now it has a very nice description. Let $E = \{ 1, e \}$ be a monoid with $1$ a unit and $e$ an idempotent. Consider it as a monoidal category with elements of $E$ as objects, no non-trivial morphisms and the monoidal product induced by the multiplication of $E$. Then your monoidal product is the Day convolution of functors $E \to \mathrm{Set}$. Since $E$ is commutative this makes $\mathrm{Set}^E$ into a closed symmetric monoidal category. It is also complete and cocomplete (even locally presentable) so all the standard theory of enriched categories is available.

    I think it is more natural (and more categorical) to think of objects in this category as pairs of sets. (Just as it is more natural to think of graded modules and rings as sequences of abelian groups as opposed to abelian groups equipped with a decomposition into a direct sum of a sequence of subgroups.)

22. • CommentRowNumber22.
    • CommentAuthorAndrew Stacey
    • CommentTimeApr 30th 2013
    • (edited Apr 30th 2013)
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexFantastic! Thanks again. Just one query: > I think it is more natural (and more categorical) to think of objects in this category as pairs of sets. What do you mean by "this category" here? Do you mean the base category that we're enriching over? If so, I agree. But you could mean the enriched category, in which case I'm confused. So we can view a cobordism category as one of these, with diffeomorphisms and cobordisms as the two pieces of the morphisms, and we can view LCTVS as one of these, with a (functorial) choice of topology on the tensor product. Then a field theory is an enriched functor from one to the other. LCTVS also has some extra structure in that there's a natural transformation from the module part to the main part. Ack, let's have some terminology! Given a category enriched over $\Set^E$ let's write $M_1(a,b)$ and $M_e(a,b)$ for the two pieces of the morphisms. So in LCTVS, we have a natural transformation $M_e(-,-) \to M_1(-,-)$. Moreover, LCTVS is actually enriched over $\LCTVS^E$ if we choose some functorial LCTVS structure on the morphism sets. Hmm, this might be a fun project to use to learn about enriched categories. Anyone feel this is doomed to failure?

    Fantastic! Thanks again. Just one query:

    I think it is more natural (and more categorical) to think of objects in this category as pairs of sets.

    What do you mean by “this category” here? Do you mean the base category that we’re enriching over? If so, I agree. But you could mean the enriched category, in which case I’m confused.

    So we can view a cobordism category as one of these, with diffeomorphisms and cobordisms as the two pieces of the morphisms, and we can view LCTVS as one of these, with a (functorial) choice of topology on the tensor product. Then a field theory is an enriched functor from one to the other.

    LCTVS also has some extra structure in that there’s a natural transformation from the module part to the main part. Ack, let’s have some terminology! Given a category enriched over $\Set^E$ let’s write $M_1(a,b)$ and $M_e(a,b)$ for the two pieces of the morphisms. So in LCTVS, we have a natural transformation $M_e(-,-) \to M_1(-,-)$. Moreover, LCTVS is actually enriched over $\LCTVS^E$ if we choose some functorial LCTVS structure on the morphism sets.

    Hmm, this might be a fun project to use to learn about enriched categories. Anyone feel this is doomed to failure?

23. • CommentRowNumber23.
    • CommentAuthorKarol Szumiło
    • CommentTimeApr 30th 2013
    • PermaLink
    Author: Karol Szumiło
    Format: MarkdownItex> What do you mean by “this category” here? Do you mean the base category that we’re enriching over? If so, I agree. Yes, that's what I mean. > So in LCTVS, we have a natural transformation $M_e(-,-) \to M_1(-,-)$. You can extend $E$ to a symmetric monoidal category $E'$ by adding an extra arrow $e \to 1$. Enrichment over $\Set^{E'}$ should capture this extra structure.

    What do you mean by “this category” here? Do you mean the base category that we’re enriching over? If so, I agree.

    Yes, that’s what I mean.

    So in LCTVS, we have a natural transformation $M_e(-,-) \to M_1(-,-)$.

    You can extend $E$ to a symmetric monoidal category $E'$ by adding an extra arrow $e \to 1$. Enrichment over $\Set^{E'}$ should capture this extra structure.

24. • CommentRowNumber24.
    • CommentAuthorAndrew Stacey
    • CommentTimeApr 30th 2013
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexRight, but then I lose the example of cobordisms. I think that LCTVS is an example of a very highly structured category (Todd Trimble and I had a long discussion about the variety of monoidal structures on it somewhere in the nCafé) so basing everything on that example might end up with something so structured that there aren't any other examples of it. On the other hand, I guess that the functor $E \to E'$ makes any $\Set^{E'}$-category automatically a $\Set^E$-category and using the $\Set^{E'}$-structure means that any properties that come from it being an enriched category (as you hint at in #21) can be assumed to respect this natural transformation. Another example would be Euclidean spaces with isometries and linear maps. That is, isometries form the $1$-morphisms and linear maps form the $e$-morphisms. (This is over $\Set^E$, of course, not $\Set^{E'}$.)

    Right, but then I lose the example of cobordisms.

    I think that LCTVS is an example of a very highly structured category (Todd Trimble and I had a long discussion about the variety of monoidal structures on it somewhere in the nCafé) so basing everything on that example might end up with something so structured that there aren’t any other examples of it.

    On the other hand, I guess that the functor $E \to E'$ makes any $\Set^{E'}$-category automatically a $\Set^E$-category and using the $\Set^{E'}$-structure means that any properties that come from it being an enriched category (as you hint at in #21) can be assumed to respect this natural transformation.

    Another example would be Euclidean spaces with isometries and linear maps. That is, isometries form the $1$-morphisms and linear maps form the $e$-morphisms. (This is over $\Set^E$, of course, not $\Set^{E'}$.)

25. • CommentRowNumber25.
    • CommentAuthorTobias Fritz
    • CommentTimeMay 1st 2013
    • (edited May 1st 2013)
    • PermaLink
    Author: Tobias Fritz
    Format: HtmlWithout having looked at any of this in detail, I wonder whether the following paper might be relevant: <a href="http://arxiv.org/abs/math/9805102">Nuclear and Trace Ideals in Tensored *-Categories</a>

    Without having looked at any of this in detail, I wonder whether the following paper might be relevant:
    
    Nuclear and Trace Ideals in Tensored *-Categories
26. • CommentRowNumber26.
    • CommentAuthoradeelkh
    • CommentTimeMay 1st 2013
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    Author: adeelkh
    Format: Markdown> (Just as it is more natural to think of graded modules and rings as sequences of abelian groups as opposed to abelian groups equipped with a decomposition into a direct sum of a sequence of subgroups.) Sorry to get off-topic, but would you mind explaining briefly why that is? I've heard someone say this before but I wasn't convinced by their argument.

    (Just as it is more natural to think of graded modules and rings as sequences of abelian groups as opposed to abelian groups equipped with a decomposition into a direct sum of a sequence of subgroups.)

    Sorry to get off-topic, but would you mind explaining briefly why that is? I've heard someone say this before but I wasn't convinced by their argument.

27. • CommentRowNumber27.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 1st 2013
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexTobias, That is indeed a very interesting paper. I read through parts of it (skipping the probabilistic bits) and there are some close connections. The main difference is that I don't want to assume that my "other morphisms" are a subset of the "true morphisms". In Hilbert spaces then this is true, but other examples fail for that. (I also *strongly* disagree with their definition of a "nuclear" object. In LCTVS then nuclear spaces do not have the property that they require. I'm pretty sure that only finite dimensional LCTVS would satisfy their definition of being a nuclear object.)

    Tobias, That is indeed a very interesting paper. I read through parts of it (skipping the probabilistic bits) and there are some close connections. The main difference is that I don’t want to assume that my “other morphisms” are a subset of the “true morphisms”. In Hilbert spaces then this is true, but other examples fail for that.

    (I also strongly disagree with their definition of a “nuclear” object. In LCTVS then nuclear spaces do not have the property that they require. I’m pretty sure that only finite dimensional LCTVS would satisfy their definition of being a nuclear object.)

28. • CommentRowNumber28.
    • CommentAuthorKarol Szumiło
    • CommentTimeMay 1st 2013
    • PermaLink
    Author: Karol Szumiło
    Format: MarkdownItex> Sorry to get off-topic, but would you mind explaining briefly why that is? I've heard someone say this before but I wasn't convinced by their argument. I'd say that this is largely a matter of aesthetics. If you like to think abstractly, then defining graded rings as sequences makes it easier to put them into a categorical context. But more importantly, if you have a graded ring (say concentrated in nonnegative degrees) defined as a sequence of abelian groups, then there is more than one way to turn it into an actual ring. The obvious thing is to take the direct sum, but you could also take the product. In fact if your ring is the cohomology ring of a space (note that the cohomology ring is first given as a sequence of cohomology groups), then you want to take the product more often than not. This is a very useful point of view for complex oriented cohomology theories. They come equipped with formal group laws which can be used to describe a lot of cohomological structure but in order to exploit them we should use products rather than direct sums (e.g. we think of cohomology of $\mathbb{C} P^\infty$ as a power series ring rather than a polynomial ring.) This point would be very opaque if we insisted on the "direct sum definition" of graded rings.

    Sorry to get off-topic, but would you mind explaining briefly why that is? I’ve heard someone say this before but I wasn’t convinced by their argument.

    I’d say that this is largely a matter of aesthetics. If you like to think abstractly, then defining graded rings as sequences makes it easier to put them into a categorical context.

    But more importantly, if you have a graded ring (say concentrated in nonnegative degrees) defined as a sequence of abelian groups, then there is more than one way to turn it into an actual ring. The obvious thing is to take the direct sum, but you could also take the product. In fact if your ring is the cohomology ring of a space (note that the cohomology ring is first given as a sequence of cohomology groups), then you want to take the product more often than not. This is a very useful point of view for complex oriented cohomology theories. They come equipped with formal group laws which can be used to describe a lot of cohomological structure but in order to exploit them we should use products rather than direct sums (e.g. we think of cohomology of $\mathbb{C} P^\infty$ as a power series ring rather than a polynomial ring.) This point would be very opaque if we insisted on the “direct sum definition” of graded rings.

29. • CommentRowNumber29.
    • CommentAuthoradeelkh
    • CommentTimeMay 1st 2013
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    Author: adeelkh
    Format: MarkdownKarol, thanks for the explanation.

    Karol, thanks for the explanation.

30. • CommentRowNumber30.
    • CommentAuthorMike Shulman
    • CommentTimeMay 2nd 2013
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    Author: Mike Shulman
    Format: MarkdownItexAdeel, my impression has been that one is not necessarily "more natural" than the other in the abstract, but that it depends on what you are doing with your graded modules. In algebraic topology and category theory, it's more natural to regard them as sequences. But I've been told that in some places at least in algebraic geometry, the other perspective is more natural.

    Adeel, my impression has been that one is not necessarily “more natural” than the other in the abstract, but that it depends on what you are doing with your graded modules. In algebraic topology and category theory, it’s more natural to regard them as sequences. But I’ve been told that in some places at least in algebraic geometry, the other perspective is more natural.

31. • CommentRowNumber31.
    • CommentAuthoradeelkh
    • CommentTimeMay 3rd 2013
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    Author: adeelkh
    Format: MarkdownThat makes sense. I can confirm that algebraic geometers use the decomposition definition, probably universally.

    That makes sense. I can confirm that algebraic geometers use the decomposition definition, probably universally.