Derivative of intersection volume Let $K$ be a convex body in $\mathbb{R}^n$ and set $f:\textrm{SL}(n)\rightarrow \mathbb{R}$ as $f(T)=\textrm{Vol}_n (TB\cap K)$ where $B$ is the Euclidean unit ball. How can we find extreme points of $f$?
What I'm looking for is some Taylor expansion of $f$, so I may write for matrices such as $Q=I_n + \epsilon F$ something in the line of $$f(Q)=f(I_n)+\epsilon f'(Q)$$
where $f'$ is a directional derivative of some sort of $f$. I believe this should amount to something like $f'(T)=\textrm{Vol}_{n-1} (\partial TB\cap K)$, but this is pure intuition, I'm not sure how this can be proven.
 A: Let us first formalise the idea of directional derivatives of matrices
Derivatives of variable matrices are usually expressed as Lie derivatives. The basic object is a Lie group, i.e., a differentiable manifold that has a group structure such that the group operations are differentiable. In our case this is the $(n-1)$-dimensional manifold $SL(n)$ consisting of all real $n\times n$ matrices with determinant $1.$ They are also precisely the linear transformations of $\mathbb R^n$ that preserve volume and orientation.
Lie theory considers $1$-parameter subgroups: differentiable homomorphisms from the simplest possible Lie group $(\mathbb R,+)$ to the Lie group under study:
$$T:\mathbb R\to SL(n):t\mapsto T_t,\hskip1cm T_{s+t}=T_sT_t.$$
The derivatives at $0$ of all such possible subgroups form the tangent space of the differentiable manifold at the unit element $T_0=I$, which in this context is called the Lie algebra. Our Lie algebra is denoted ${\mathfrak{sl}(n)}$ and it consists of all $n\times n$ matrices with trace $0.$
The one-parameter group generated by a matrix $A\in\mathfrak{sl}(n)$ is given by the exponential mapping
$$\exp:\mathfrak{sl}(n)\to SL(n):A\mapsto\exp(A)=\sum_{i=0}^\infty\frac{A^i}{i!}$$
which answers your question about a power series expansion.
The derivative of $\textrm{Vol}_n (T_tB\cap K)$ is more easily evaluated if we replace the indicator functions of the compact sets $B$ and $K$ with differentiable functions $\phi$ and $\psi$ that approximate them. So we are looking at the quantity
$$V_t=\int_{\mathbb R^n}\phi(T_t^{-1}x)\psi(x)ds$$
Let us evaluate the derivative of $V_t$ at $t=0.$
$$\eqalign{
\frac{dV_t}{dt}(t=0)&=\frac{d}{dt}\int_{\mathbb R^n}\phi(T_t^{-1}x)\psi(x)dx\\
&=\int_{\mathbb R^n}\frac{d\phi(T_t^{-1}x)}{dt}\psi(x)dx\\
&=\int_{\mathbb R^n}\nabla\phi\cdot (-A)x\psi(x)dx\\
}$$
As $\phi$ approaches the indicator of $K$ its gradient converges to a distribution that is concentrated on $\partial K$ and models the inward normal $(-n)$ of $K$ (since $K$ is convex its boundary has an inward normal almost everywhere). Thus we have
$$\eqalign{
\frac{dV_t}{dt}(t=0)&=\int_{\partial K\cap B}Ax\cdot n\ dS\\
}$$
Alternatively, notice that $Ax$ is a divergence-free vector field (because the trace of $A$ is $0$) so the integral is also equal to
$$\eqalign{
\frac{dV_t}{dt}(t=0)&=\int_{\partial B\cap K}Ax\cdot n\ dS\\
}$$
(the reason why these two integrals do not have opposite signs, as one would expect from partial integration, is that the interpretation of $n\ dS$ as an outward normal vector is different according to whether the 'outward' means out of $K$ or out of $B$)
The second integral is different from your intuitive idea but there is a close resemblance.
Higher derivatives of $V_t$ are not guaranteed to exist without additional conditions on the shape of $K.$ This can be intuitively understood by noticing that the first derivative is an integral where not only the integrand, but also the area of integration depends on $t.$ In fact the first derivative need not be a continuous function as can be seen in $2$ dimensions by letting $B=B(c=(5;0),r=1),$ $K$ the upper half of $B$ and $A=\left(\begin{matrix}0&1\\-1&0\end{matrix}\right)$ (generator of the rotations around the origin). 
