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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?

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up vote 7 down vote accepted

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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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. –  Yuji Tachikawa Sep 26 '13 at 10:25
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