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.



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> The question is whether the round sphere is the only surface with this property.
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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 $t=0$?



area variation of a closed surface under ${\rm SL}(3)$
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](https://en.wikipedia.org/wiki/Ellipsoid#Surface_area). 


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. 


