Let $Ss(4m)$ be the $Z/2Z$ quotient of $Spin(4m)$ which is not $SO(4m)$. (This group is somtimes called the semi-spin group.) Its $Z/2Z$ cohomology was determined e.g. by Baum and Browder MR article. Is the $Z/2Z$ cohomology of its classifying space determined somewhere? What is \begin{equation} H^*(BSs(4m),Z/2Z) ? \end{equation}

Update: In the string theory application I have in mind, it would be enough to know it up to degree 11. Does this make the determination any easier?


Tetsu Nishimoto kindly performed the computation, and allowed me to reproduce it here. --Yuji

Proposition: The mod-2 cohomology $H^*(BSs(16m);\mathbb Z/2)$ of the classifying space of the Lie group $Ss(16m)$, is isomorphic to the following algebra up degree $ \leq 11$: $$ \mathbb Z/2[x_2, x_3, x_5, x_9, y_4, y_6, y_7, y_{10}, y_{11}] /(x_2y_7+x_3y_6+x_5y_4+x_2x_3y_4). $$ Within $\ast \leq 11$, the action of $Sq^k$ is given by $$ \begin{array}{|c|c|c|c|c|c|} \hline & Sq^1 & Sq^2 & Sq^3 & Sq^4 & Sq^5 \\ \hline x_2 & x_3 & x_2^2 & & & \\ x_3 & 0 & x_5 & x_3^2 & & \\ x_5 & x_3^2 & 0 & 0 & x_9 & x_5^2 \\ x_9 & x_5^2 & 0 & & & \\ y_4 & 0 & y_6 & y_7 & y_4^2 & \\ y_6 & y_7 & 0 & 0 & y_{10} & y_{11} \\ y_7 & 0 & 0 & 0 & y_{11} & \\ y_{10} & y_{11} & & & & \\ \hline \end{array}. $$

Let us describe the outline of the computation. First we need to quote the structure of $H^*(Ss(16m);\mathbb Z/2)$ as a Hopf algebra from Proposition 4.1 of

Hopf Algebra Structure of mod 2 Cohomology of Simple Lie groups
K. Ishitoya, A. Kono and H. Toda
Publ. RIMS, Kyoto Univ. 12 (1976) 141-167 electric version

The proposition states that, in the range $\ast \leq 10$, $H^*(Ss(16m);\mathbb Z/2)$ is isomorphic as an algebra to $$ \Delta (w_3, w_5, w_6, w_7, w_9, w_{10}) \otimes \mathbb Z/2[\bar{v}] $$ where $\deg w_i = i$, $\deg \bar{v} = 1$. The generators other than $w_7$ are primitive, while the coproduct of $w_7$ is given by $$ \bar\psi (w_7) = \bar{v} \otimes w_6 + \bar{v}^2 \otimes w_5 + \bar{v}^4 \otimes w_3. $$ The action of $Sq^k$ within the range $\ast \leq 10$ is given by $$ \begin{array}{|c|c|c|c|c|c|c|c|} \hline & Sq^1 & Sq^2 & Sq^3 & Sq^4 & Sq^5 \\ \hline w_3 & 0 & w_5 & w_6 = w_3^2 & & \\ w_5 & w_6 & 0 & 0 & w_9 & w_{10} = w_5^2 \\ w_6 & 0 & 0 & 0 & w_{10} & \\ w_7 & m\bar{v}^8 & w_9 & w_{10} & & \\ w_9 & w_{10} & & & & \\ \hline \end{array}. $$

Next, we consider the Rothenberg-Steenrod spectral sequence $$ E_2 = \mathrm{Cotor}_{H^*(Ss(16m);\mathbb Z/2)} (\mathbb Z/2, \mathbb Z/2) \Longrightarrow H^*(BSs(16m);\mathbb Z/2). $$

We first need to compute the $E_2$ term. Here we use May's spectral sequence $$ E'_1=\mathrm{Cotor}_{A'}(k,k) \Longrightarrow \mathrm{Cotor}_{A}(k,k). $$ Here, $A'$ is a Hopf algebra such that it is isomorphic as an algebra with $A'$ such that every generator is primitive. When the characteristic of $k$ is $2$, $\mathrm{Cotor}_{A'}(k,k)$ is a polynomial ring whose generators are in one-to-one correspondence with the primitive elements of $A'$. Here we take $k=\mathbb{Z}/2$ and $A=H^*(Ss(16m);\mathbb Z/2)$. Then, up to degree 11, we have $$ \mathrm{Cotor}_{A'}(k,k) = \mathbb{Z}/2[[v],[v^2],[v^4],[v^8],[w_3],[w_5],[w_6],[w_7],[w_9],[w_{10}]] $$ The differential at $E_2$ is given by $$ d_2([w_7]) = [v][w_6]+[v^2][w_5]. $$ Note that from the construction of May's spectral sequence the term $[v^4][w_3]$ vanish. As all the other differentials are zero, $E'_\infty$ up to degree 11 is isomorphic to $$ \mathbb Z/2[[v],[v^2],[v^4],[v^8],[w_3],[w_5],[w_6],[w_9],[w_{10}]]/([v][w_6]+[v^2][w_5]). $$ It is easily seen that in $\mathrm{Cotor}_A(k,k)$ the relation corresponding to $[v][w_6]+[v^2][w_5]$ is $[v][w_6]+[v^2][w_5]+[v^4][w_3]$. Therefore $\mathrm{Cotor}_{H^*(Ss(16m);\mathbb Z/2)} (\mathbb Z/2, \mathbb Z/2) $ is given by $$ \mathbb Z/2 [[\bar{v}], [\bar{v}^2], [\bar{v}^4], [\bar{v}^8], [w_3], [w_5], [w_6], [w_9], [w_{10}]]/ ([\bar{v}][w_6]+[\bar{v}^2][w_5]+[\bar{v}^4][w_3]) $$ up to degree 11 as algebras.

We note here that $[\bar{v}^{2^j}] \in E_2^{1,2^j}$, $[w_i] \in E_2^{1,i}$, and that these generators all correspond to primitive elements. When the degrees are higher this is not necessarily the case. The relation came from the coproduct of $w_7$, as we saw above.

The differentials are given in the range $r \geq 2$ as \begin{align*} & d_r : E_r^{1,1} \longrightarrow E_r^{1+r,1-(r-1)} = E_2^{1+r,1-(r-1)} = 0, \\ & d_r : E_r^{1,3} \longrightarrow E_r^{1+r,3-(r-1)} = E_2^{1+r,3-(r-1)} = 0. \end{align*} Therefore, $[\bar{v}]$ and $[w_3]$ are permanent cycles. In general, when $x$ is a permanent cycle, for any cohomology operation $\theta$ $\theta x$ is also a permanent cycle. Other generators can be written as \begin{align*} & [\bar{v}^2] = Sq^1[\bar{v}], \quad [\bar{v}^4] = Sq^2Sq^1[\bar{v}], \quad [\bar{v}^8] = Sq^4Sq^2Sq^1[\bar{v}], \\ & [w_5] = Sq^2[w_3], \quad [w_6] = Sq^3[w_3], \quad [w_9] = Sq^4Sq^2[w_3], \quad [w_{10}] = Sq^5Sq^2[w_3] \end{align*} and therefore these are also permanent cycles. Therefore, up to degree 11, we have $E_{\infty} = E_2$.

Let us now define elements of $H^*(BSs(16m);\mathbb Z/2)$. Let $x_2$ be a representative of $[\bar{v}]$, and $y_4$ be a representative of $[w_3]$. $x_2$ is uniquely determined but there are two elements $y_4$ and $y_4+x_2^2$ representing $[w_3]$. This freedom is used below when we fix the relations. Let us further set \begin{align*} & x_3 = Sq^1 x_2, \quad x_5 = Sq^2 x_3, \quad x_9 = Sq^4 x_5, \\ & y_6 = Sq^2 y_4, \quad y_7 = Sq^1 y_6, \quad y_{10} = Sq^4 y_6, \quad y_{11} = Sq^1 y_{10} \end{align*} then they are representatives of $[\bar{v}^2]$, $[\bar{v}^4]$, $[\bar{v}^8]$, $[w_5]$, $[w_6]$, $[w_9]$, $[w_{10}]$, respectively. Using the Adem relation, we can determine how $Sq^k$ acts on these elements, giving the table shown above.

Finally let us determine the relation in $H^*(BSs(16m),\mathbb Z/2)$ corresponding to the relation $[\bar{v}][w_6]+[\bar{v}^2][w_5]+[\bar{v}^4][w_3]$ of $\mathrm{Cotor}_{H^*(Ss(16m);\mathbb Z/2)} (\mathbb Z/2, \mathbb Z/2)$. When $k \geq 3$, the basis of $E_2^{k,9-k}$ can be given by $$ x_2^2x_5, \quad x_3^3, \quad x_2^3x_3, \quad x_2x_3y_4 $$ and therefore the degree-9 relation $r$ in $H^*(BSs(16m);\mathbb Z/2)$ can be given by $$ r = x_2y_7 + x_3y_6 + x_5y_4 + a_1x_2^2x_5 + a_2x_3^3 + a_3x_2^3x_3 + a_4x_2x_3y_4 \quad (a_i \in \mathbb Z/2). $$ From $$ Sq^1 r = (1+a_4)x_3^2y_4 + (a_1+a_3)x_2^2x_3^2 $$ we have $a_4 = 1$ and $a_1 = a_3$. From \begin{align*} Sq^2 r & = x_2^2y_7 + (a_1+a_2)x_3^2x_5 + a_1x_2^4x_3 + a_1x_2x_3^3 + a_1x_2^3x_5 + x_2^2x_3y_4 + x_2x_5y_4 + x_2x_3y_6 \\ & = (a_1+a_2)x_3^2x_5 + (a_1+a_2)x_2x_3^3 \end{align*} we have $a_1 = a_2$. Then the relation is given by $$ r = x_2y_7 + x_3(y_6+a_1x_3^2) + (x_5+x_2x_3)(y_4+a_1x_2^2). $$ When $a_1 = 1$, we exchange $y_4$ by $y_4+x_2^2$. Then $y_6$ is exchanged with $y_6+x_3^2$ and $y_{10}$ is exchanged with $y_{10}+x_5^2$, while all the other generators are fixed. The relation then becomes $$ r = x_2y_7 + x_3y_6 + (x_5+x_2x_3)y_4 $$ and has the same form as the relation for the case $a_1 = 0$. This completes the determination of the relation.

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I don't think the cohomology is known. As an indication of this I would refer to the 2007 paper "On the Stiefel–Whitney classes of the representations associated with Spin(15)" by Mamoru Mimura and Tetsu Nishimoto (see http://www.msp.warwick.ac.uk/gtm/2007/11/p007.xhtml). There they state that "calculating $H^\ast(BSs(16);Z/2Z)$ seems to be as difficult as calculating $H^\ast(BE_8;Z/2Z)$" (which is also still unknown).

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  • $\begingroup$ Thank you very much for the reference. I added an update in the question, saying that I only want to know the structure up to degree 11. $\endgroup$ – Yuji Tachikawa Sep 26 '13 at 10:25

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