0

As in symmetric space $Sym^{g}(\Sigma)$, $g\geq 2$, $\Delta$ is defined to be the diagonal space , i.e, for any element $x=(x_{1}, X_{2},...,x_{g})\in \Delta$, there are existing $x_{i}=x_{j}$, where $i\neq j$. So my question is why $\Delta$ is a codimension-2 manifold?
would you please write it down if no mind? Thank you?

flag
1 
 I feel like this is more appropriate for math.SE. – Marco Golla Oct 19 2011 at 8:59
I agree with Marco. Anyway, think of the easiest case $g=2$: then $\Delta$ is naturally isomorphic to $\Sigma$, so it is a divisor in $Sym^g(\Sigma)$. This means that the real codimension of $\Delta$ in this case is $2$.Now you should be able to generalize this argument for higher values of $g$. – Francesco Polizzi Oct 19 2011 at 9:05
I agree with Marco too, but anyway, maybe you should think about $g=2$ and $\Sigma$ is a point or an open interval. – Allen Knutson Oct 19 2011 at 9:24
3 
 Be careful: $\Delta$ has (real) codimension $2$ (assuming $\Sigma$ is a Riemann surface), but it isn't typically a manifold. It is usually singular along the locus of collections supported at fewer than $g-1$ points. – Jack Huizenga Oct 19 2011 at 13:30

closed as too localized by Mark Grant, Denis Serre, Tim Perutz, Andreas Blass, Ryan Budney Oct 19 2011 at 13:25

Browse other questions tagged or ask your own question.