Questions about the Moduli Space of Vector Bundles of rank 2 with trivial determinant Let $M_0$ be the moduli space of rank 2 semi-stable vector bundles over X with trivial determinant which is a singular projective variety of dimension $3g-3$. $M_0$ is constructed as the $GIT$ quotient of the $R(2,p)$ by $SL(p)$. 
Let $M_0^{s}$ be the points of moduli space corresponding to the semi-stable points of $R(2,p)$. Then the singular locus of $M_0$ is the Kummer Variety $\mathbb{K}$ or the complement of $M_0^{s}$. The elements of $\mathbb{K}$ are of the form $L\oplus L^{-1}$, where L is a line bundle of degree 0. $\mathbb{K}_0$ be bundles of the form $L\oplus L^{-1}$ such that $L=L^{-1}$, i.e., $L^2$ is trivial. 
Question 1- What is meant by a Kummer Variety? Why $\mathbb{K}_0$ is the Kummer Variety of dimension g?
$\mathbb{Z_2}$ acts on $Jac_0$ by involution i.e., $(0,L)\longmapsto L$ and $(1,L)\longmapsto L^{-1}$
$\mathbb{K}$ $\cong$ $Jac_0//\mathbb{Z_2}$. There are $2^{2g}$ number of fixed points $\mathbb{Z_2^{2g}}=\{[L\oplus L^{-1}:L\cong L^{-1}]\}$of the action.
Question2.- Is this isomorphism is isomorphism as varieties? 
Question3- Why are there are $2^{2g}$ number of fixed points of the action. 
Thus $M_0$ has a stratification
$M_0=M_0^{s}\coprod(\mathbb{K}-\mathbb{Z_2^{2g}})\coprod Z_2^{2g}$. 
Over each point in the deepest strata $\mathbb{Z_2^{2g}}$ there is unique closed orbit in $R(2,p)^{ss}$. By deformation theory, the normal space of the orbit at a point h, which represents $L\oplus L^{-1}$ where $L\cong L^{-1}$ is, 
$H^{1}(End_0(L\oplus L))\cong H^{1}(\mathcal(O))\oplus \mathcal{sl}(2)$
Question4- Why is the last statement true?
 A: I can give a partial answer to this question.
If you're looking at rank two vector bundles of the form $L \oplus L^{-1}$ over a curve $X$ of genus $g$ where $L$ has degree zero, then this is going to be isomorphic to $Pic_0(X)/(\mathbb{Z}/2)$, where the map in one direction is the obvious map
$$
L \mapsto [L \oplus L^{-1}]
$$
The point is that you can't tell the difference between $L \oplus L^{-1}$ and $L^{-1} \oplus L$, since if $L$ is a line bundle of degree zero (i.e. an element of $Pic_0(X)$), then so is $L^{-1}$.
Anyhow, $Pic_0(X)$ is an abelian variety of dimension $g$, and its quotient is called the Kummer variety, which is also of dimension $g$. This can be seen in a number of different ways, including the fact that $Pic_0(X) \cong H^1(X, \mathcal{O}_X)/H^1(X,\mathbb{Z})$. Since $H^1(X,\mathcal{O}_X)$ is a $g$-dimensional vector space and the image of $H^1(X,\mathbb{Z})$ has maximal rank, the quotient is topologically isomorphic to $S^1 \times \cdots \times S^1$ (where there are $2g$ factors).
The fixed points of the action of $\mathbb{Z}/2$ are clearly those line bundles $L$ such that the swapping of factors $L \oplus L^{-1} \leftrightarrow L^{-1}\oplus L$ makes no difference; these are exactly those lines bundles $L$ such that $L^2 = \mathcal{O}_X$ as suggested, which are precisely the 2-torsion points of $Pic_0(X)$; as this is $2g$ copies of $S^1$, there are $2^{2g}$ of these.
I see no reason why this would not be an isomorphism of varieties. The action of $\mathbb{Z}/2$ on $Pic_0(X)$ is reasonably well-behaved, so I see no reason why they aren't isomorphic...
Anyhow, one concrete example to think of is in the case $g = 2$. In such a case we get as the Kummer variety the quotient of an abelian surface $A$ by the action of $\pm1$, which has 16 fixed points which locally look like $\mathbb{C}^2/\pm1$. This is singular, and we can resolve each of these with a single blowup to obtain an exceptional $\mathbb{P}^1$; the resulting surface is a smooth K3 surface call the Kummer surface.
