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In Stiefel-Whitney classes of real representations of finite groups, J. Algebra 126 (1989), no. 2, 327–347, Gunawardena, Kahn and Thomas dealt with the question, whether the cohomology ring $H^*(G, \mathbb{Z}/2)$ was generated as an abelian group on the Stiefel-Whitney Classes of flat vector bundles over BG.

The outcome is that the cohomology ring of a family of split metacyclic groups $$G_{m,n}^{+}= \langle t, s \mid t^{2^m +1}=s^{2^n}=1, \, sts^{-1} =t^{{2^m}+1}\rangle $$

with the property that a map onto $D_{2^m}$ has the subgroup $\langle t\rangle$ in its kernel, for specific choice of parameters $1+m-n\neq m$ has a third cohomology class which is not the Stiefel-Whitney class of any vector bundle.

Are there examples of finite groups, where the second and first cohomology groups are not generated by Stiefel-Whitney classes ?

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    $\begingroup$ Certainly not first cohomology, which classifies real line bundles by their first Stiefel-Whitney class. I wasn't able to conclude anything about $w_2$; one might try to analyze the universal case; that is, see if the map $w_2: BSO \to B(\Bbb Z/2, 2)$ has a section. I wasn't able to run the obstruction theory argument (there are infinitely many obstructions...) $\endgroup$
    – mme
    Feb 12, 2019 at 5:16
  • $\begingroup$ Has anyone thought of an example with a finite group, since this question was asked? I really wonder if a finite 2-group could have $H^2$ not generated by Stiefel-Whitney classes -- seems unlikely. $\endgroup$
    – Pierre
    Nov 26, 2021 at 14:45

2 Answers 2

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$\mathrm{H}^1(G,\mathbb{Z}/2) \cong \hom(G,\mathbb{Z}/2)$ consists precisely of first Stiefel–Whitney classes.

So your question is to decide whether there is a finite group with an element in $\mathrm{H}^2(G,\mathbb{Z}/2)$, i.e. with a double cover, which is not a second Stiefel–Whitney class, i.e. which is not pulled back from the Spin double cover of an oriented real representation. (What about $w_1$s? Well, first, if $A$ and $B$ are line-bundles, then $w_2(A\oplus B) = w_1(A)w_1(B)$. Second, for $V$ arbitrary, $V \oplus 3\det(V)$ is orientable and $w_2(V \oplus 3\det(V)) = w_2(V)$.)

I don't know the answer, but I can answer in the negative the version where $G$ is a Lie group. Namely, consider the double cover $\mathrm{SO}(8) \to \mathrm{PSO}(8)$. I claim that this double cover is not pulled back from any Spin double cover of any representation $V$ of $G = \mathrm{PSO}(8)$. Equivalently, I claim that the class in $\mathrm{H}^2(B\mathrm{PSO}(8), \mathbb Z/2)$ classifying this double cover is not a Stiefel–Whitney class.

It suffices to restrict the map $\mathrm{PSO}(8) \to \mathrm{SO}(V)$ to the maximal torus $T \subset \mathrm{PSO}(8)$, where I can calculate with weights. The root lattice $\Lambda$ of $T$ is the sublattice $\Lambda \subset \mathbb Z^4$ consisting of vectors with even dot product with $(1,1,1,1)$. The roots are $(\pm1,\pm1,0,0)$, $(\pm1,0,\pm1,0)$, $(\pm1,0,0,\pm1)$, $(0,\pm1,\pm1,0)$, $(0,\pm1,0,\pm1)$, and $(0,0,\pm1,\pm1)$. The weights of the vector representation of $\mathrm{SO}(8)$ are $(\pm1,0,0,0)$, $(0,\pm1,0,0)$, $(0,0,\pm1,0)$, and $(0,0,0,\pm1)$. The Weyl group is $W = 2^3{:}4!$, where the $2^3$ normal subgroup acts as reflections in an even number of coordinates, and $4! = S_4$ acts as permutations.

I will use only that $V|_T$ is a sum of Weyl-invariant representations of $T$. I.e. I claim for any Weyl-invarnat representation $T \to \mathrm{SO}(V)$, the Spin double cover of $\mathrm{SO}(V)$ trivializes when restricted to $T$.

An irreducible representation $V$ of $T.W$, i.e. a Weyl-invariant irrep of $T$, consists of a single Weyl orbit through some vector $\lambda \in \Lambda$. Note that $-1 \in W$. It follows that $V$ is the underlying real representation of a complex representation $U$ of $T$. (Emphasis: this holds over $T$, but not over $T.W$ or $\mathrm{PSO}(8)$.) For instance, you can take $U$ to consist of the positive weights appearing in $V$. (The positive weights are those for which the first nonzero entry is positive.)

It is a general fact that if $V$ is the underlying real representation of a complex representation $U$, then $w_2(V) = c_1(U) \mod 2$. But $c_1(U)$ is exactly the sum of weights appearing in $U$. And my claim that $w_2(V) = 0$ is equivalent to claiming that $\frac12 c_1(U) \in \Lambda$, which is to say I need to show that $\frac12 c_1(U) \in \mathbb Z^4$ and that $\langle \frac12 c_1(U), (1,1,1,1) \rangle \in 2\mathbb Z$.

I'll let the highest weight of $V$ be $\lambda = (a,b,c,d) \in \Lambda$, so that $a+b+c+d \in 2\mathbb Z$ (and remember that my $V$ is simply the Weyl-orbit through that highest weight). The positive Weyl chamber is $a \geq b \geq c \geq d \geq -c$, so that's where $\lambda$ lives. The "reflection" outer automorphism sends $(a,b,c,d) \mapsto (a,b,c,-d)$, so I can assume $d \geq 0$.

I'll casebash the calculation.

Case 1: $a>0$, $b=c=d=0$. The weights of $V$ are $(\pm a, 0,0,0), (0, \pm a,0,0)$, $(0,0,\pm a,0)$, $(0,0,0,\pm a)$, so the weights of $U$ are $(a,0,0,0)$, $(0,a,0,0)$, $(0,0,a,0)$, $(0,0,0,a)$, and $c_1(U) = (a,a,a,a)$. Note that $a \in 2\mathbb Z$ by assumption, and so $\frac12 c_1(U) \in \mathbb Z^4$. Furthermore $\frac12 \langle c_1(U),(1,1,1,1)\rangle = 2a$ is even (in fact divisible by $4$).

Case 2: $a \geq b >0$, $c = d = 0$. If $a=b$, then the weights of $U$ are $(a, \pm a, 0,0)$, $(a, 0, \pm a, 0)$, $(a, 0,0\pm a, )$, $(0,a,\pm a,0)$, $(0,a,0,\pm a)$, $(0,0,a,\pm a)$ and $\frac12 c_1(U) = (3a,2a,a,0) \in \mathbb Z^4$ and $3+2+1 = 6$ is even. If $a > b$, then $\frac12 c_1(U) = (3(a+b),2(a+b),a+b,0)$. In either case, we get a term divisible by $6$ for each permutation of the $(a,b)$.

Case 3: $c>0$, $d=0$. The weights of $U$ come in four sets: $(+x, \pm y,\pm z,0)$, $(+x, \pm y, 0, \pm z)$, $(+x, 0, \pm y, \pm z)$, and $(0, +x, \pm y, \pm z)$, where $(x,y,z)$ ranges over the permutations of $(a,b,c)$. Then $\frac12 c_1(U)$ is a sum over permutations of $(a,b,c)$ of a term like $(6x, 2x, 0,0)$, manifestly integral and with integer dot-product with $(1,1,1,1)$.

Case 4: If $d>0$, then the weights of $U$ are the vectors of the form $$ (+w, +x, +y, +z), (+w, +x, -y, -z), (+w, -x, +y, -z), (+w, -x, -y, +z)$$ where $(w,x,y,z)$ ranges over the permutations of $(a,b,c,d)$. Then $\frac12 c_1$ is a sum of terms like $(2w,0,0,0)$.

Note that the only case where $\frac12 c_1(U)$ came close to leaving $\mathbb Z^4$ was Case 1, where I needed that $a \in 2\mathbb Z$, and the only case where $\langle \frac12 c_1(U), (1,1,1,1)\rangle$ came close to leaving $2\mathbb Z$ was Case 2.

I remark that the triality outer automorphism relates $\lambda = (2,0,0,0)$ (Case 1) to $\lambda = (1,1,1,1)$ (Case 4). For the former, $\frac12 c_1(U) = (1,1,1,1)$, and for the latter $\frac12 c_1(U) = (2,0,0,0)$.

If you repeat the argument for a general PSO group, I think you will find that all real representations of $\mathrm{PSO}(n)$ are Spin (and so you get no nontrivial double covers from them) whenever $n$ is divisible by $8$.

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    $\begingroup$ Can you extend this argument to showing that there is no map $BPSO(8) \to BSO$ which pulls back $w_2$ to this class? That would answer the 'universal' question I posed in the comments. (The weaker question, whether or not there is a section of $BSO(n) \to K(\Bbb Z/2, 2)$, is easier to solve: the first has cohomology with polynomial growth, and the second has cohomology with superpolynomial growth.) $\endgroup$
    – mme
    Feb 12, 2019 at 19:03
  • $\begingroup$ @MikeMiller I think you are asking subtlety that I am not sensitive to. I believe I showed that for any map $BPSO(8) \to BSO$, the restriction of $w_2$ vanishes. But now I'm hesitant about the difference between representations and maps to classifying spaces and... $\endgroup$ Feb 12, 2019 at 19:31
  • $\begingroup$ @TheoJohnson-Freyd If I understand correctly, you showed that any map $BPSO(8)→BSO(n)$ vanishes for every $n$. This is not quite the same as showing that all maps $BPSO(8)→BSO$ vanish, since $BPSO(8)$ is not a (retract of a) finite CW-complex and so a priori you could have a map to $BSO$ that does not factor through any finite $n$ $\endgroup$ Feb 12, 2019 at 19:51
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    $\begingroup$ @DenisNardin Ah, got it. Well, I don't know the answer, then. $\endgroup$ Feb 12, 2019 at 23:04
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I recently wondered about this question myself. For anyone else who might have wondered about this, I'm going to share my findings.

First, I would like to note that the paper of Gunarwardena, Kahn and Thomas mentioned in the post actually contains examples of metacyclic groups with a second cohomology class that's not a Stiefel-Whitney class. In the proof of their Theorem 2 (bottom of page 336 to top of following page) they show that there are type 1 cases where $x$ (in degree 2) is not in the subring of Stiefel-Whitney classes. One can understand the reason for this in the specific example case of the group number 27 of order 64 in GAP's SmallGroups library. This group has only four real representations, all of which are one-dimensional, all other representations are complex. So the second Stiefel-Whitney classes that can appear are actually first Chern classes of complex representations, and therefore integral. But then, in the Serre spectral sequence for the extension of cyclic by cyclic, the way the cyclic quotient acts on the cyclic subgroup is such that there are degree 2 classes on the subgroup which are not fixed integrally, but are fixed mod 2 by the quotient group. Such a class will give rise to a mod 2 cohomology class which is not the reduction of an integral class, and so not a Stiefel-Whitney class.

Second, I would like to point to a great resource on the relation between mod 2 cohomology and the subring generated by Stiefel-Whitney classes. Pierre Guillot's webpage https://irma.math.unistra.fr/~guillot/research/cohomology_of_groups/index.html has a listing of some groups of small order with presentations of their cohomology rings in terms of Stiefel-Whitney classes. There are some examples where a degree 3 class is not in the subring generated by Stiefel-Whitney classes (e.g. the groups [16,6], [32,15], [32,17], [32,37], [64,31], [64,40], [64,45] and some more of order 64, all with GAP's SmallGroups library numbering). Unfortunately no degree 2 examples.

So I wanted to understand the situation better. For an initial sweep, one could imagine doing the following: Since computing Stiefel-Whitney classes for a finite group $G$ is difficult, let's go for something simpler, and consider an elementary abelian subgroup $E\subset G$. Any representation of $G$ restricted to $E$ splits as direct sum of one-dimensional ones, so it's easy to compute the restriction of the Stiefel-Whitney classes of the representation to $E$. If in degree $n$, the rank of the subring of $H^*(E,\mathbb{F}_2)$ generated by restrictions of Stiefel-Whitney classes is smaller than the rank of the restriction map $H^*(G,\mathbb{F}_2)\to H^*(E,\mathbb{F}_2)$ in cohomology, then there is a difference between cohomology ring and Stiefel-Whitney subring already for $G$. This slightly naive procedure can be implemented in GAP, and it produces a number of groups where the second cohomology is not generated by Stiefel-Whitney classes:

  • in order 32, the groups with SmallGroup numbers 4 and 13.
  • in order 64, the groups with SmallGroup numbers 3, 10, 14, 15, 27, 35, 48, 57, 62, 63, 64, 74, 78, 80, 81, 84, 106, 113, 123, 157, 162.

Some of them look metacylic (and number 27 is), but some of them have elementary abelian subgroups of ranks 3 and 4. Many more examples for order 128. It seems there are a number of examples, but not so many that one would bump into them all the time.

This procedure also finds over a 100 groups of order 64 (that's roughly 1/3 of the total number) where the third cohomology is not generated by Stiefel-Whitney classes, complementing Pierre Guillot's collection somewhat. Note that the procedure might miss examples where the difference between cohomology and the Stiefel-Whitney subring is not visible on maximal elementary abelian subgroups. (There are examples like this in Guillot's collection, such as [16,6], [32,37], or [64,45].) So it appears that there are some examples where the difference between cohomology and Stiefel-Whitney subring is nilpotent. Extending the procedure to find these cases would require a more refined approach, actually computing the Stiefel-Whitney classes in group cohomology of $G$ directly, maybe using explicit Brauer induction or stuff like that. Also, of course, there may be bugs in my implementation of the naive search, or the underlying software packages used (mainly GAP and HAP), but I did check a small number of examples by hand, sort of: the character splitting and characteristic class computations can be done by hand or using GAP, and for the restriction maps on cohomology one can use the database https://users.fmi.uni-jena.de/cohomology/ with the output of the p-group-cohomology package of Simon King and David Green.

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