# Example of a finite group $G$ with low dimensional cohomology not generated by Stiefel-Whitney classes of flat vector bundles over $BG$

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 ?

• 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...) – Mike Miller Feb 12 at 5:16

$$\mathrm{H}^1(G,\mathbb{Z}/2) \cong \hom(G,\mathbb{Z}/2)$$ 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 $$8$$.

• 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.) – Mike Miller Feb 12 at 19:03
• @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... – Theo Johnson-Freyd Feb 12 at 19:31
• @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 $n$ – Denis Nardin Feb 12 at 19:51
• @DenisNardin Ah, got it. Well, I don't know the answer, then. – Theo Johnson-Freyd Feb 12 at 23:04