Hyperelliptic loci in Teichmueller spaces Let ${\cal M}_g$ be the moduli space of smooth complex genus $g$ curves, let ${\cal H}_g\subset {\cal M}_g$ be the hyperelliptic locus and set ${{\cal H}}'_g$ to be the preimage of ${\cal H}_g$ in the Teichmueller space.
While working on a problem I arrive at two results that can't be reconciled unless ${\cal H}'_3$ is disconnected.
While it seems a bit strange to me that ${\cal H}'_3$ should be disconnnected, I don't see why it should't be. So I'd like to ask whether this is known or known to be false.
[sorry, had to cut this into small paragraphs, otherwise the tex part wouldn't show properly.]
 A: That $\mathcal H'_g$ can't be connected for $g \geq 3$, isn't this just the "lift to Teichmuller space" of the result that the mapping class group isn't the hyperelliptic group?
In particular, the path components of $\mathcal H'_g$ are indexed by the cosets of the hyperelliptic group in the mapping class group.  No? 
edit: Anweshi, I'm not sure how you're thinking about Teichmuller space but the answer to your question can be seen in many ways, you don't have to use the  language of orbifolds, it's just a convienient container.  In my mind I suppose I think of a path in Teichmuller space as a motion of the surface -- make this concrete using Fenchel-Nielsen coordinates, for example.  So if you have a path that connects one point to another there is an associated diffeomorphism of the surfaces that stretches/twists the metric appropriately and matches up the markings of the surfaces.  So if you go between two points in your $\mathcal H_g'$ covering the same point in $\mathcal H_g$ the relating diffeomorphism is in the hyperelliptic group (since the hyperelliptic group is a subgroup of the mapping class group).  This is how you `see' the cosets of the hyperelliptic group in the mapping class group as indexing $\pi_0 \mathcal H'_g$. 
A: There's a slight issue I believe with the other answers. If we consider moduli space as an orbifold (of complex dimension $3g-3$), and the hyperelliptic locus an immersed suborbifold (of complex dimension $2g-1$ or so), then we may (essentially) identify the hyperelliptic locus with the orbifold of $2g+2$ points on $S^2$, obtained by quotienting each Riemann surface by the hyperelliptic involution. However, how does one know that this space doesn't "cross" itself? Imagine by analogy an immersed geodesic curve on a hyperbolic surface, such that each complementary component is a disk: the preimage in the universal cover is connected, when taken as a union of geodesics, even though each geodesic lift is embedded. 
This sort of crossing does not occur for the hyperelliptic locus. If two branches of the hyperelliptic universal cover in Teichmuller space were to intersect, then there would be a single Riemann surface fixed by two distinct hyperelliptic involutions. But a hyperelliptic involution fixes precisely the $2g+2$ Weierstrauss points of the surface, and is therefore uniquely determined, a contradiction. So in fact the hyperelliptic locus is "embedded", in the sense that each lift corresponds to a fixed set of a hyperelliptic involution, and distinct hyperelliptic involutions give distinct components in Teichmuller space. 
A: Teichmuller space is the universal cover of $M_g$.  Thus, if $X$ is a locus in $M_g$, the preimage of $X$ in Teichmuller space is connected if and only if the induced map
$$\pi_1(X) \to \pi_1(M_g)$$
is surjective.  In your case, you are asking:  is the hyperelliptic mapping class group in genus $3$ the whole of the genus $3$ mapping class group $\Gamma_3$?  No, it isn't:  the hyperelliptic mapping class group is the centralizer of an involution in $\Gamma_3$.
