Let $(A,L)$ be a polarized abelian variety. I know that the degree of the polarization is the Euler characteristic of $L$, so that
$d = \chi(L) = \dim H^0(A,L)$
since $L$ is ample.


I've read in a lot of papers the sentence



> 
> Let $(A,L)$ be a polarized abelian variety of dimension $g$ and of type $(d\_1, \dots, d\_g)$, etc.
> 
> 
> 


I've seen the formal definition of type in Birkenhake-Lange book, but I cannot relate this definition with the degree of a polarization, although I am pretty sure they are related.


Indeed, when a statement says "type $(1,1,\dots,1,d)$", I think of "degree $d$", but it is just an intuition.


So my question is:



> 
> what is exactly a type of an abelian variety? How do we relate type and degree?
> 
> 
> 


Thanks for help!



Type vs degree of a polarized abelian variety
**To answer your first question**, i.e., what is type on a (polarized) abelian variety $A$? It is already answered [here](https://mathoverflow.net/q/177246). Roughly speaking, for a polarized abelian variety $(A,L)$, there is an integral basis $\{dx\_i,dy\_i\}\_{i=1}^{g}$ of $H^1(A,\mathbb Z)$, and positive integers $d\_1,\ldots, d\_g$ with $d\_i|d\_{i+1}$ such that
$$c\_1(L)=\sum\_i d\_idx\_i\wedge dy\_i.\tag{1}\label{1}$$
The $g$-tuple $(d\_1,\ldots,d\_g)$ is called the type of $(A,L)$.


For example, when $A=J(C)$ is the Jacobian variety of a smooth projective curve $C$, one can choose $L$ to be principal polarization ($d\_1=\cdots=d\_g=1$). This follows from the fact that there is a symplectic basis $\{\gamma\_i,\delta\_i\}\_{i=1}^g$ on $H\_1(J(C),\mathbb Z)=H\_1(C,\mathbb Z)$ such that $\gamma\_i\cdot\delta\_j=\delta\_{ij}$ and $\gamma\_i\cdot\gamma\_j=\delta\_i\cdot\delta\_j=0$.


**For your second question**, in fact, the book (Birkenhake-Lange, p.119) defines the *degree of a polarization* of $L$ to be the product $\prod\_id\_i$. So the relation you're looking for is probably


$$\chi(L)=\prod\_id\_i.\tag{2}\label{2}$$
*Proof of \eqref{2}.* First, by Hirzeburch-Riemann-Roch theorem, one obtains
$$\chi(L)=\int\_Ach(L)td(A)=\int\_Ach(L)=\int\_A\frac{c\_1(L)^g}{g!},\tag{3}\label{3}$$


where the second and the third identities follow from the fact that the tangent bundle of $A$ is trivial and the definition of the Chern character on a line bundle.


Now, by \eqref{1}, one has
$$c\_1(L)^g=(g!\prod\_id\_i)dx\_1\wedge dy\_1\wedge\cdots\wedge dx\_g\wedge dy\_g,$$ so $\int\_Ac\_1(L)^g=g!\prod\_id\_i$. Combine with identity \eqref{3}, the equality \eqref{2} follows. $\Box$


As a final remark, the proof of the identity \eqref{2} can be more concrete: By Kodaira vanishing theorem, all higher degree cohomology of $L$ vanish, so $\chi(L)=\dim H^0(A,L)$, one it reduces to show that there are exactly $\prod\_id\_i$ independent global sections on $L$. These sections correspond to theta functions on $A$. One can refer to Griffiths-Harris, p.317-320 for how these theta functions are found explicitly.


