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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeOct 4th 2013

added a table with some homotopy groups in the unstable range to orthogonal group – Homotopy groups

• CommentRowNumber2.
• CommentAuthorDavidRoberts
• CommentTimeNov 27th 2014

Extended by two rows. $SO(6)$ is the highest rank in which peculiar things happen below dimension 10. I think it would be good to extend it up to $\pi_11$ or $\pi_12$. Mimura and Mimura-Toda have this information for the last two rows. I’m looking around to see if I can find it for the other groups.

• CommentRowNumber3.
• CommentAuthorDavidRoberts
• CommentTimeNov 27th 2014
• (edited Nov 27th 2014)

Ah, the venerable Encyclopedic Dictionary of Mathematics has a nice big table of these homotopy groups, for all the groups of interest. Also, the Abanov document has evaporated off the web.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeNov 27th 2014

Is there a way to see this online? In GoogleBooks somewhere?

• CommentRowNumber5.
• CommentAuthorDavidRoberts
• CommentTimeNov 27th 2014
• (edited Nov 27th 2014)

Not that I’m aware of, which is a pity.

I’m sticking the table in, at least up to $\pi_12(SO(12))$, since this probably covers all the orthogonal/homotopy groups of interest to string theory. I can do more if requested; the table goes up to $\pi_15(SO(17))$.

• CommentRowNumber6.
• CommentAuthorDavidRoberts
• CommentTimeNov 27th 2014

In fact, in writing this out, it becomes clear that something different is going to happen for Ninebrane, since stably ($SO(11)$ and beyond, hence in the right range for 11d supergravity) we have nontrivial torsion homotopy groups in dimension 8 and 9. $SO(11)$ also has a nonstable $\mathbb{Z}/2$ in $\pi_10$, and its $\pi_11$, though $\mathbb{Z}$, is not in the stable range. Hmm

• CommentRowNumber7.
• CommentAuthorDavidRoberts
• CommentTimeNov 27th 2014

One can see a copy of the table in the EDM in page 970 of the Handbook of Algebraic Topology This chapter, number 19, was written by Mimura.

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeNov 27th 2014

Thanks for expanding the table so much!

Yes, ninebrane structure is meant to kill just the fractional $p_3$ of a Fivebrane structure, not the two $\mathbb{Z}/2\mathbb{Z}$-copies in between.

The way this appears in string theory is not actually that the structure group needs to be highly connected as such (that happens to be a byproduct), it’s that the polynomials in the $p_i$ appear as anomalies, and these need to be removed.

• CommentRowNumber9.
• CommentAuthorDavidRoberts
• CommentTimeNov 27th 2014

But how does one get the fractional $p_3$? Does this exist even if the structure group doesn’t lift through those two copies?

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeNov 27th 2014

There is a $\mathbb{Z}/2\mathbb[p_3$ on $B \mathrm{Fivebrane}$ pulled back from $B \mathrm{String}$ and it will be a multiple of something. Do you mean how to pick the specific fraction?

The way I suggested to do it is just to let the process of higher Lie integration run. There is that $L_\infty$-cocycle $\mu_{12}$ on $\mathfrak{fivebrane}$ and just integrating that as in Cech Cocycles for Differential characteristic Classes (schreiber) gives some $\mathbf{B} \mathrm{Ninebrane}$ whose geometric realization is a candidate $B \mathrm{Ninebrane}$. I don’t know what the fraction comes out as, but the way the Lie integration works one is guaranteed that one kills a minimal element.

• CommentRowNumber11.
• CommentAuthorDavidRoberts
• CommentTimeNov 27th 2014

Ah, so the fact it’s a Lie algebra theoretic fact means you don’t see the $\mathbb{Z}/2$? I worry though that in doing the Lie integration one needs to pick at some point a filler of some simplex. But let me remind myself how it works, first, before complaining it doesn’t! :-)

OK, I see. It’s just a formal thing, but one can then ’go down’ the Whitehead tower and construct the quotients by various to get the intermediate stages. The Lie integration just skips them altogether.

One point: under Lie integration#integrating_the_string_lie_2algebra_to_the_string_lie_2group there is what I think is a typo: $\mathbf{B}^n\mathbb[D^2_\ast \to D$ should probably by $D^2_\ast \to G$, but I hesitate to change in case I’ve missed something.

• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeNov 27th 2014

Yes! Absolutely.

And yes, that should be a typo. If you have a second, please fix it. Myself I need to rush off..

• CommentRowNumber13.
• CommentAuthorDavidRoberts
• CommentTimeNov 27th 2014

Will do.

• CommentRowNumber14.
• CommentAuthorDavidRoberts
• CommentTimeMar 24th 2015

Fixed two errors in the table in the $SO(6)$ row.

• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeMar 24th 2015

thanks. I wonder what happened there. Did I just make a misprint while copying? Probably.

• CommentRowNumber16.
• CommentAuthorDavidRoberts
• CommentTimeMar 24th 2015

@Urs, no it was probably me, since I added the citation for the $SO(6)$ row via the exceptional isomorphism.

• CommentRowNumber17.
• CommentAuthorUrs
• CommentTimeMay 3rd 2016
• (edited May 3rd 2016)

I have added to orthogonal group:

• statement and proof that $O(n) \hookrightarrow O(k)$ is an $(n-1)$-equivalence (here)

• statement and proof that the Stiefel manifold $V_n(k)$ is $(n-1)$-connected (here).

The latter I am also copying over to Stiefel manifold

• CommentRowNumber18.
• CommentAuthorUrs
• CommentTimeMay 4th 2016

I have now added the analogous statements with their analogous proofs also to unitary group.

• CommentRowNumber19.
• CommentAuthorDavidRoberts
• CommentTimeNov 28th 2021
• (edited Nov 28th 2021)

Added more precise reference for the result that the induced map $\mathbb{Z}^2=\pi_3(SO(4)) \to \pi_3(SO(5))=\mathbb{Z}$ is $(a,b)\mapsto 2a+b$ in

• Itiro Tamura, On Pontrjagin classes of homotopy types of manifolds, Journal of the mathematical society of Japan, Vol. 9 No. 2 , 1957 pdf

For reference, $\mathbb{Z} =\pi_3(SO(3)) \to \pi_3(SO(4))=\mathbb{Z}^2$ is $x\mapsto (x,0)$. The same maps will work for the groups $Spin(n)$ for $n=3,4,5$, which is what I care about.

I wanted this, so I could check what the map $H^4(BSp(2),\mathbb{Z})\to H^4(BSp(1)\times BSp(1),\mathbb{Z})$ is.

Urs has been very careful with the isomorphism $H^4(BSpin(4),\mathbb{Z})\to H^4(BSp(1)\times BSp(1),\mathbb{Z})$, making clear what the map $\mathbb{Z}^2 = H^4(BSpin(4),\mathbb{Z})\to H^4(BSpin(3),\mathbb{Z})=\mathbb{Z}$ should be, namely projection on the first factor. This agrees with the homotopy group computation, since we can use the isomorphisms $\pi_3(G)^*\simeq \pi_4(BG)^*\simeq H_4(BG,\mathbb{Z})^*\simeq H^4(BG,\mathbb{Z})$ for a simply-connected compact Lie group $G$.

However, it is less clear to me (and probably only because I haven’t thought about it hard enough) how to reconcile $\pi_3(Spin(4)) \to \pi_3(Spin(5))$, or equivalently, $H_4(BSpin(4),\mathbb{Z})\to H_4(BSpin(5),\mathbb{Z})$, with the explicit computation changing basis at Spin(4)#euler_class_and_pontryagin_class and the detail at Spin(5)#cohomology. The notation with the various references to $\frac1 2 p_1$ and $\frac1 4 p_1$ is obscuring things for me, so I don’t know how I should think of the map

$H^4(BSpin(5),\mathbb{Z}) \to H^4(BSpin(4),\mathbb{Z})$ $\mathbb{Z}[\tfrac1 2 p_1]^* \to \mathbb{Z}[\tfrac1 4 p_1\otimes 1,1\otimes \tfrac1 4 p_1]^*$

as written in this basis.

• CommentRowNumber20.
• CommentAuthorDavidRoberts
• CommentTimeNov 28th 2021

So I can see that the dual of $H_4(BSpin(4),\mathbb{Z})\to H_4(BSpin(5),\mathbb{Z})$, using the basis described by Tamura (chased through the isomorphisms) should be represented by the matrix

$\binom 2 1\colon \mathbb{Z} = H^4(BSpin(5),\mathbb{Z}) \to H^4(BSpin(4),\mathbb{Z}) \simeq H^4(BSp(1)\times BSp(1),\mathbb{Z}) = \mathbb{Z}^2$

so at least on the first factor I get that we should have $\frac1 2 p_1 \mapsto 2 \cdot (\frac 1 4 p_1 \otimes 1)$—the notation is cute and makes sense. But I can’t see what’s going on with the second factor, yet. I mean, we could just have $\frac1 2 p_1 \mapsto (1\otimes \frac 1 4 p_1)$ (or equivalently, $\frac{1}{2}\big( \chi+ \frac{1}{2}p_1 \big)$), and the fact this is not a stable class means that any notational niceties around $p_1$ are probably not really relevant here.

The reason I’m thinking about this, is to try to understand better some of the deep detail of the big diagram from Twisted Cohomotopy implies twisted String structure on M5-branes, in particular the structure of the $Sp(2)$-equivariant quaternionic Hopf fibration, in its model as

$\Big(Sp(2)/Sp(1)\Big)//Sp(2) \longrightarrow \Big(Sp(2)/(Sp(1)\times Sp(1))\Big)//Sp(2)$
• CommentRowNumber21.
• CommentAuthorUrs
• CommentTimeNov 28th 2021

No time right now and in the next days (this would have been good to discuss earlier!).

Just to note that $\tfrac{1}{4}p_1$ is an integral cohomology generator on $B Spin(3)$, e.g. Lem. 2.1 in Čadek & Vanžura 1998 , which you probably have seen.

• CommentRowNumber22.
• CommentAuthorDavidRoberts
• CommentTimeNov 28th 2021
• (edited Nov 28th 2021)

I think I may have figured it out. There’s a change of basis going on so the obvious basis for $\pi_3(Sp(1)\times Sp(1))$ is not what Tamura (following Steenrod’s book) is using. His first basis vector is (1,1) using the canonical product group structure, the second is (1,0). The first corresponds to the image of the generator of $\pi_3(Spin(3))$ under the canonical $Spin(3)\to Spin(4)$. So using the obvious basis coming from $\pi_3(Sp(1)\times Sp(1))$, the map induced on $\pi_3$ by $Sp(1)\times Sp(1)\to Sp(2)$ really is the sum map. After changing basis, it gives the above result.

• CommentRowNumber23.
• CommentAuthorDavidRoberts
• CommentTimeMar 20th 2022

This is just a reminder for when editing functionality comes back, to write up a small explanation about the subtlety around the choice of bases as discussed in #19, #20 and #22.

• CommentRowNumber24.
• CommentAuthorDavidRoberts
• CommentTimeMar 21st 2022
• (edited Mar 21st 2022)

Want to record another point: Tamura uses the basis $(1,1), (1,0)$ relative to the obvious Künneth decomposition of $H^4$ of $Spin(4)\simeq Spin(3)\times Spin(3)$ (which should be the same as the product structure coming from $\pi_3$ of the same.

But Čadek & Vanžura use the basis $(1,1), (0,1)$, as recorded by Urs, under the isomorphism just mentioned. This helps resolve my query in #20, where I was struggling with understanding the use of $1\otimes \frac{1}{4} p_1$.