The full braid group on $n$ strands $B\_n$ admits a surjective homomorphism $p\colon\thinspace B\_n\to \Sigma\_n$ onto the symmetric group on $n$ letters, which takes a braid to the induced permutation of its ends. The kernel $P\_n$ is well understood; it is the pure braid group on $n$ strands.


What about $p^{-1}(A\_n)$, where $A\_n$ is the alternating group on $n$ letters? Let me call this group $E\_n$ for now, because I think it should be called the *even braid group*. However an internet search using this name (and others such as "orientation preserving braids", "positive braids" and so on) came up blank.



> 
> Does this group $E\_n$ have a name, and has it been studied anywhere in the literature? 
> 
> 
> 


I would be particularly interested in computations of the cohomology rings of these groups.




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**Update:** 
It occurred to me that there was one obvious name I hadn't searched for, which was "alternating subgroups of braid groups". This led me to the following preprint,


<http://arxiv.org/abs/1207.3947>


which has a section on finding presentations for these groups (the alternating subgroup of the braid group associated to a Coxeter system $(G,S)$ is denoted $\mathcal{B}^+(G)$ in Section 5). So it seems that they do indeed appear in the literature, although not until surprisingly recently!



Does this subgroup of "even braids" have a name?
I don't know if these groups have been studied before, but I can say something about their cohomology rings, at least over $\mathbb{Q}$. Namely, we have $H^k(E\_n;\mathbb{Q}) = \mathbb{Q}$ if $k=0,1$ and $H^k(E\_n;\mathbb{Q}) = 0$ for $k \geq 2$. Of course, this is the same as the cohomology of the ordinary braid group as computed by Arnold in


V. I. Arnold, On some topological invariants of algebraic functions, Trudy Moscov. Mat. Obshch. 21 (1970), 27-46 (Russian), English transl. in Trans. Moscow Math. Soc. 21 (1970), 30-52.


Recall that if $H$ is a finite-index normal subgroup of $G$, then $G$ acts on $H^k(H;\mathbb{Q})$ and using the transfer map we have that $H^k(G;\mathbb{Q})$ is equal to the invariants of this action.


For braid groups, the action of $B\_n$ on $H^k(PB\_n;\mathbb{Q})$ factors through an action of the symmetric group $S\_n$, so $H^k(PB\_n;\mathbb{Q})$ is a *representation* of $S\_n$ and $H^k(B\_n;\mathbb{Q})$ is the trivial subrepresentation $\{\text{$v \in H^k(PB\_n;\mathbb{Q})$ $|$ $\sigma v = v$ for all $\sigma \in S\_n$}\}$.


Let's now consider $E\_n$. In this case, the above argument shows that $H^k(E\_n;\mathbb{Q})$ is the subrepresentation $\{\text{$v \in H^k(PB\_n;\mathbb{Q})$ $|$ $\sigma v = v$ for all $\sigma \in A\_n$}\}$. 


Now, representations of finite groups over $\mathbb{Q}$ decompose into direct sums of irreducible representations. The only two irreducible representations of $S\_n$ that restrict to the identity on $A\_n$ are the trivial representation and the alternating representation. As we said above, the trivial representation corresponds to $H^k(B\_n;\mathbb{Q})$, so we conclude that
$$H^k(E\_n;\mathbb{Q}) = W \oplus H^k(B\_n;\mathbb{Q}),$$
where $W \subset H^k(PB\_n;\mathbb{Q})$ is the direct sum of all alternating subrepresentations.


The above calculation is thus equivalent to the assertion that the alternating representation does not occur in $H^k(PB\_n;\mathbb{Q})$. This follows from the calculation of $H^k(PB\_n;\mathbb{Q})$ as a representation of $S\_n$ which was done by in the paper "Coxeter group actions on the complement of hyperplanes and special involutions" by Felder-Velesov; see [here](http://arxiv.org/abs/math/0311190).




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The above ref to Felder-Velesov was suggested by Vladimir Dotsenko; I originally included the argument below, which only works for $n \gg k$.


It's quite hard to decompose $H^k(PB\_n;\mathbb{Q})$ into irreducibles; however the paper "Representation Theory and Homological Stability" (see [here](http://arxiv.org/abs/1008.1368)) by Church and Farb introduces a recipe that they call "representation stability" which describes how the decomposition of $H^k(PB\_{n+1};\mathbb{Q})$ into irreducibles can be constructed from the decomposition of $H^k(PB\_n;\mathbb{Q})$ into irreducibles, at least for $n$ large. Their results are hard to summarize briefly, but they do imply that the alternating representation does not occur (it is not "stable" in their sense), again at least for $n$ large.


