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Tian has defined bisectional curvature for unit and perpendicular tangent vectors x,y , as follow
R(x,y,x,y)+R(x,Jy,x,Jy). If bisectional curvature be constant, is there any relationship between sectional curvature and bisectional curvature? In this case (bisectional curvature is constant) what we can say about conjugate points of the manifold?

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The bisectional curvature (on unit vectors) is never constant, see Robert Bryant's first comment to your previous question mathoverflow.net/questions/105687/… –  YangMills Sep 9 '12 at 23:52
    
Tian in "canonical metrics in kahler geometry" has proved that bisectional curvature of $\mathbb{C}^n$, $\mathbb{B}^n$ and $\mathbb{CP}^n$ are constant. Robert Bryant in his comment talked about holomorphic bisectional curvature. –  Reza Sep 10 '12 at 5:28
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You shouldn't trust books blindly, and try to compute by yourself the bisectional curvature of $\mathbb{CP}^n$, and you will see that it is not constant. If you normalize the Fubini-Study metric so that it has constant holomorphic sectional curvature $1$, then the bisectional curvatures will vary between $1/2$ and $1$ (again, see Robert Bryant's comment). –  YangMills Sep 10 '12 at 16:43
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up vote 9 down vote accepted

Since you seem to have some basic misunderstandings about these concepts, I will try to clarify the definitions.

Let $(M,J)$ be a complex manifold, with a Riemannian metric $g$ which is Hermitian, i.e. $g(JX,JY)=g(X,Y)$ for all $X,Y\in TM$ real tangent vectors. Assume furthermore that $g$ is Kähler, i.e. that the real $2$-form $\omega(X,Y)=g(JX,Y)$ is closed.

Then it is well-known that the Riemann curvature tensor of $g$ satisfies the "Kähler identity" $$R(X,Y,Z,W)=R(X,Y,JZ,JW).$$ We will use the convention that if $X,Y$ are nonparallel unit vectors, then the sectional curvature of the $2$-plane they span is $R(X,Y,Y,X)$.

Then for a unit vector $X$ we define its holomorphic sectional curvature $H(X)$ to be the sectional curvature of the $2$-plane spanned by $X$ and $JX$, i.e. $$H(X)=R(X,JX,JX,X).$$ In analogy with this, we define the bisectional curvature of two unit vectors $X,Y$ to be $$R(X,JX,JY,Y).$$ The name comes from the fact that (using the Bianchi and Kähler identities) $$R(X,JX,JY,Y)=R(X,Y,Y,X)+R(X,JY,JY,X),$$ so that if $X$ and $Y$ are not parallel it is indeed equal to the sum of two sectional curvatures (incidentally, this might be the relationship you wanted between sectional and bisectional curvature).

Then every simply connected Kähler manifold with constant holomorphic sectional curvature is holomorphically isometric to one of the three model spaces $\mathbb{C}^n, B^n$ and $\mathbb{CP}^n$ with (a rescaling of) their standard metrics. This is proved for example in Kobayashi-Nomizu "Foundations of Differential Geometry, Vol.2", Theorem 7.9.

In fact, by Proposition 7.3 in that same book, $H(X)$ is constant equal to $c$ iff $$R(X,Y,Z,W)=\frac{c}{4}\bigg(g(X,W)g(Y,Z)-g(X,Z)g(Y,W)+g(X,JW)g(Y,JZ)$$ $$-g(X,JZ)g(Y,JW)+2g(X,JY)g(W,JZ)\bigg).$$ In particular, in this case you have that the bisectional curvature of $X,Y$ equals $$R(X,JX,JY,Y)=\frac{c}{2}\bigg(1+g(X,Y)^2+g(X,JY)^2\bigg).$$

As you can see, this expression is not constant (as you vary $X,Y$ among unit vectors), but it varies between $\frac{c}{2}$ and $c$ according to the relative position of the $2$-planes spanned by $(X,JX)$ and $(Y,JY)$.

In particular you see that there is no such thing as a non-flat "Kähler manifold with constant bisectional curvature", because if there was such a thing, in particular the holomorphic sectional curvature would be constant, but then the curvature tensor would be given by the above formula, and the bisectional curvature would actually be NOT constant. On the other hand, if $c=0$, then all the curvatures are constant because the metric is flat.

It goes without saying that Tian's book has a recurrent typo, and wherever he says "constant bisectional curvature" you should substitute "constant holomorphic sectional curvature".

Apart from the book of Kobayashi-Nomizu, you can also read this paper of Goldberg-Kobayashi.

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Of course, it leaves flat spaces that do have constant bisectional curvature, right? –  Benoît Kloeckner Sep 12 '12 at 8:39
    
Yes, of course, sorry about that. I will edit my answer accordingly. –  YangMills Sep 12 '12 at 14:23
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