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](http://www.ams.org/mathscinet-getitem?mr=189063) [article](http://www.sciencedirect.com/science/article/pii/0040938365900017). 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?



Cohomology of the classifying space of $Ss(4m)$
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](https://www.jstage.jst.go.jp/article/kyotoms1969/12/1/12_1_141/_article)
> 
> 
> 


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.


