area variation of a closed surface under ${\rm SL}(3)$ Let $\Sigma$ a closed oriented embedded surface in $R^3$. When $\Sigma$ is a round sphere, then for any smooth curve $A(t) \in SL(3)$ through the identity (i.e. $A(t) \in R^{3\times 3}$, $\det(A(t))=1$, $A(0)=I$), we have
$$
\frac{d}{dt}\bigg|_{t=0} {\rm Area} (A(t) \Sigma) = 0.
$$ 
This can be proved either by a direct calculation, or a quite simple argument based on the isoperimetric inequality.

The question is whether the round sphere is the only surface with this property.

Put differently: If $\Sigma$ is not a round sphere, is there always a direction $v \in sl(3)$ such that the surface area of $\exp(vt) \Sigma$ has a nonzero variation at $t=0$?
 A: In fact, for any surface $\Sigma$, there exists a matrix $A_0 \in SL_3(\mathbb{R})$ so that $A_0\cdot\Sigma$ is a critical point for variations by $sl(3)$. The point is that if we look at the orbit $\{A\cdot\Sigma, A\in SL_3(\mathbb{R})\}$, the area function  $\Phi: SL_3(\mathbb{R})\to \mathbb{R}$ given by $\Phi(A)=Area(A\cdot\Sigma)$ is proper on this orbit. That is, as $A \to \infty$, $\Phi(A)\to \infty$. Granted this, then there is an absolute minium of the function $\Phi$ at some matrix $A_0\in SL_3(\mathbb{R})$. Hence $A_0\cdot \Sigma$ will be a critical point for $\Phi(A(t))$ for any path $A(t)\in SL_3(\mathbb{R})$, $A(0)=A_0$. 
To see the claim, first note that it is true for $\Sigma_r=S^2_r$, a round sphere of radius $r$. The orbit $A\cdot S^2_r$ consists of ellipsoids of axes $a,b, c $ such that $abc=r^3$. Then the area of these ellipsoids $\to \infty$ if (say) $a \to \infty$, by an approximate area formula for ellipsoids.  
For the general case, assume that $\Sigma$ is an embedded closed surface containing the sphere $S^2_r$ bounding the ball $B_r$. For any $A\in SL_3(\mathbb{R})$, consider the nearest point projection $\rho_{r,A}: \mathbb{R}^3 \to A\cdot B_r$. Then this map is area decreasing, since the derivative is a linear projection at each point in $\mathbb{R}^3-A\cdot B_r$. Hence $Area(A\cdot \Sigma) \geq Area(\rho_{r,A} A\cdot \Sigma) \geq Area(A\cdot S^2_r)$. So the area function $\Phi$ is proper since it is for ellipsoids. 
A: If $\Sigma$ enclosed domain $D_0$, then we divide $D$ into small
cubes. If $x,\ y,\ z,\ w$ are vertices in $D$, then $$ {\rm vol}\
D={\rm det}\ [x-w\ y-w\ z-w]$$ so that $$ {\rm vol}\ AD={\rm vol}\
D$$
Hence ${\rm vol}\ D_0={\rm vol}\ AD_0$. By isoperimetric inequality,
${\rm area}\ \Sigma$ is smallest when $\Sigma$ is a sphere.
