Generalized Birman exact sequence for surfaces with boundaries

Question

Let $S_g^n$ be a surface of genus g with n boundaries and let $Mod(S_g^n)$ be its mapping class group. We will also denote by $S_{g,m}^n$ a surface of genus g with n boundaries and m punctures.

The classical Birman exact sequence is

$$ 1 \to \pi_1(UT(S_g^{n-1}) \to Mod(S_g^n) \to Mod(S_g^{n-1}) \to 1$$

where $UT(S_g^{n-1})$ is the unitary tangent bundle over $S_g^{n-1}$ and the last map corresponds to capping one of the boundaries by a disk.

I am wondering what is the kernel of the map $f \colon Mod(S_g^n) \to Mod(S_g^{n-k})$ obtained by capping $k$ boundaries by $k$ disks.

It is a known fact, proven by Birman, that the kernel of the map $Mod(S_{g,m}) \to Mod(S_{g,m-k})$ obtained by forgetting $k$ punctures can be identified with $B_k(S_{g,m})$, the braid group on $k$ strands over $S_{g,m}$.

Therefore, if $PB_k(S_{g,m})$ is the pure braid group on $k$ strands over $S_{g,m}$, my best guess is that the kernel of $f$ is isomorphic to $PB_k(S_{g,m}) \oplus \mathbb{Z}^k$, where $\mathbb{Z}^k$ corresponds to the free abelian group with generators the Dehn twists around the boundaries we capped. Indeed, we can factor $f$ into $f = f_1f_2$ where $f_2$ caps the boundaries by punctures disks and $f_1$ forgets the punctures.

Is this correct? If so, is there another expression for this kernel? (Similarly to how $\pi_1(S_g^{n-1}) \oplus \mathbb{Z}$ can be identified with $\pi_1(UT(S_g^{n-1})$).

Answer 1

It's almost right, your guess.

The easiest description is geometric, generalizing Birman's description $\pi_1(UT(S_g^{n-1}))$. You look instead at the configuration space parametrizing $k$ distinct points in $S_g^{n-k}$, together with unit tangent vectors at each of these $k$ points. So this is an open subspace of the cartesian product $UT(S_g^{n-k})^k$. The fundamental group of this space is the group you care about. It is sometimes called the framed braid group.

The framed braid group is an extension of the pure braid group $PB_k(S^m_g)$ by $\mathbf Z^k$, as one sees geometrically by fibering the framed configuration space over the usual configuration space with fiber $(S^1)^k$. For an open surface the tangent bundle is trivial, which implies that this $(S^1)^k$-bundle is trivial, and thus the extension is trivial, too. In this case your guess is correct. But for closed surfaces you get a nontrivial central extension.