Another proof of the bidisc and the ball are biholomorphically inequivalent? Does this outline of a proof work?
Consider the ball and the bidisc in $\mathbb{C}^2$.  Give each space its Bergman metric.  To show that the ball and the bidisc are not holomorphic, it is enough to show that they are not isometric. 
One way to distinguish the two spaces is their sectional curvature.  I think I have shown that the sectional curvature of the Bergman metric of the ball is constant and negative, whereas the sectional curvature of the bidisc is nonpositive non constant. For example in the plane generated by the vectors $\langle 1,0 \rangle$ and $\langle 0,1 \rangle$ the section curvature is 0, but in the plane generated by $\langle 1,0\rangle$ and $\langle i,0 \rangle $ the sectional curvature is negative.
Is this true?  Is there anything subtle I might have missed?  I have seen a lot of pretty convoluted proofs of this fact, and I would think that this basic outline would be recorded in print somewhere if it is true, but I cannot seem to find it.
 A: In general, the ball $B^n$ with the Bergmann metric is isometric to the Hermitian symmetric space $SU(1,n)/S(U(1)\times U(n))$ where $SU(1,n)$ denotes the special pseudo-unitary group 
with respect to the indefinite Hermitian inner product with signature $(1,n)$ and $S(U(1)\times SU(n))$ denotes the subgroup of block matrices of sizes $1\times 1$ and $n\times n$. Note that this subgroup is the fixed point set of the involution of $SU(1,n)$
given by conjugation by the diagonal matrix with entries $(-1,1,\ldots,1)$. 
For an arbitrary Hermitian symmetric space given as $G/K$ where $G$ is connected and $K$ 
is a symmetric subgroup (i.e. open subgroup in the fixed point set of an invoutive automorphism of $G$), the identity components of the isometry group and the group of holomorphic automorphisms both coincide with $G$. Hence in the case of $B^n$ this group is $SU(1,n)$. 
Next, the $n$-polydisc is the product $B^1\times\cdots\times B^1$ ($n$-times) and it
carries the structure of a Hermitian symmetric space, product of $n$ copies of 
$SU(1,1)/S(U(1)\times U(1))\cong SL(2,\mathbb R)/SO(2)$, the unit disk in $\mathbb C$. 
As such, the identity component of its group of holomorphic automorphism is $SU(1,1)\times\cdots SU(1,1)$ ($n$ copies). It is readily seen that this group is not isomorphic to $SU(1,n)$ if $n>1$ (for instance, they have different dimensions, $3n$ and $n^2+2n$).  
A: You can look at the holomorphic sectional curvature. For the ball it is constant. Actually the constancy of the holomorphic sectional curvature of the Bergman metric distinguishes the ball (and domains biholomorphic to it) by an old theorem of Lu Qi Keng. 
