# 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.

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Sectional curvature of the complex ball is negative but not constant. Otherwise, everything is correct. The most difficult part is to show that the Riemannian metrics you have are indeed Bergman. –  Misha Jan 30 '13 at 4:54

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 $n^2+2n$).

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Thanks - this is really great. Do you have a reference for this? It doesn't directly address the curvature question. I am still reasonably convinced that I didn't goof too badly in my computation, and the the sectional curvatures alone distinguish the two spaces. –  Steven Gubkin Feb 18 '13 at 22:19
All of this should be in the book by Helgason on Symmetric spaces. In particular, the relation between holomorphic automorphisms and isometries is Lemma 4.3, ch. VIII. The relation between bounded symmetric domains and Hermitian symmetric spaces is sect. 7 of same chapter. Of course, this is very general and you may want to see things more directly in your case. Sectional curvature of symmetric spaces is easy to calculate. In case of the ball, it lies between $-4$ and $-1$ (depending on the normalization). In case of the polydisc, you have planes of curvature zero as it is a Riem. product. –  Claudio Gorodski Feb 19 '13 at 14:39
The fact that $SU(1,1)^n$ is the group of holomorphic automorphisms of the $n$-polydisc is an easy application of Schwarz Lemma. Ditto the group of holomorphic automorphisms of the unit disc in ${\mathbb C}^n$. –  Venkataramana Mar 5 '13 at 2:47

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.

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IIRC this result is only true for smoothly bounded domains, so it does not apply to the bidisc. Isn't the holomorphic sectional curvature of the bidisc also constant? –  Steven Gubkin Feb 18 '13 at 14:17
Actually it is true just when the Bergman metric is complete (some add that also the domain must be bounded but this can be changed to any condition yielding that the Bergman kernel and metric are well defined) so it works also for the bidisc. Some version of Lu's theorem works also for manifolds, see the recent book "The Geometry of Complex Domains" by Greene-Kim-Krantz –  ahmed sulejmani Feb 18 '13 at 14:59