Martin's Blog

Rosati involutions

Posted by Martin Orr on Wednesday, 05 December 2012 at 16:04

I intend to return to the basic theory of abelian varieties and write write a few posts on their endomorphism algebras and associated moduli spaces. To begin with, I will discuss the Rosati involution which is an involution of the endomorphism algebra coming from a polarisation. The existence of such an involution is crucial for the classification of endomorphism algebras which I will discuss next.

Involutions and bilinear forms

To motivate the definition of the Rosati involution, we will begin by considering the correspondence between bilinear forms on a vector space and anti-automorphisms of its endomorphism algebra. This correspondence is also used in the classification of algebras with involution.

Let k be a field and V a finite-dimensional k-vector space. If we have a nonsingular bilinear form \psi on V, then we for each linear map f \colon V \to V, there is a unique adjoint map f^\dag satisfying  \psi(fx, y) = \psi(x, f^\dag y) \text{ for all } x, y \in V. We can construct f^\dag explicitly by letting \lambda_\psi be the map V \to V^\vee sending y to \psi(-, y) and then  f^\dag = \lambda_\psi^{-1} \circ f^\vee \circ \lambda_\psi.

This defines an anti-automorphism of the matrix algebra \operatorname{End}_k V: that is, a linear bijection \operatorname{End}_k V \to \operatorname{End}_k V which reverses the order of multiplication. For example, if V = k^n and \psi is the standard symmetric bilinear form then \dag \colon \mathrm{M}_n(k) \to \mathrm{M}_n(k) is matrix transposition.

Theorem. The map sending a bilinear form to its adjoint anti-automorphism is a bijection between

  1. equivalence classes of nonsingular bilinear forms on V, up to multiplication by scalars; and
  2. anti-automorphisms of \operatorname{End}_k V.

The main thing to be proved is that every anti-automorphism comes from some bilinear form. This relies on the Skolem-Noether theorem, that every automorphism of \mathrm{M}_n(k) is conjugation by some invertible matrix.

We say that an anti-automorphism is an involution if its square is the identity. Under the bijection, a bilinear form corresponds to an involution if and only if it is either symmetric or alternating.

The Rosati involution

Let A be an abelian variety. Recall that a polarisation is an isogeny \phi \colon A \to A^\vee associated with an ample line bundle. This should be thought of as being "like a bilinear form on A" although no such thing exists directly. Being "associated with a line bundle" implies the symmetry condition \phi = \phi^\vee, and the ampleness is a positivity condition.

We can therefore apply the same formula as in linear algebra to define an involution of the endomorphism algebra:  \alpha^\dag = \phi^{-1} \circ \alpha^\vee \circ \phi \text{ for } \alpha \in \operatorname{End} A \otimes \mathbb{Q} where \phi^{-1} is the inverse of \phi in \operatorname{Hom}(A^\vee, A) \otimes \mathbb{Q}. This is called the Rosati involution associated with the polarisation \phi. (Note that the involution depends on the choice of polarisation. It preserves the order \operatorname{End} A if the polarisation is principal, but not necessarily otherwise.)

If we look at homology, either the \ell-adic Tate modules over any field or the H_1-Hodge structure over the complex numbers, then polarisations give rise to symplectic forms (Weil pairings or Riemann forms). The Rosati involution satisfies the adjoint identity  \psi(\alpha_* x, y) = \psi(x, \alpha_*^\dag y) for these symplectic forms, and also for the Hermitian form on the tangent space of a complex abelian variety.

Positivity of the Rosati involution

Let E = \operatorname{End} A. For x \in E, we define \operatorname{Tr}_{E/\mathbb{Q}}(x) to be the trace of the \mathbb{Q}-linear map y \mapsto xy \colon E \to E.

The Rosati involution is a positive involution, meaning that  \operatorname{Tr}_{E/\mathbb{Q}}(\alpha \alpha^\dag) > 0 \text{ for all } \alpha \in E - \{ 0 \}. The justification for this definition is that an involution of \mathrm{M}_n(\mathbb{R}) is positive if and only if its corresponding bilinear form on \mathbb{R}^n is either positive or negative definite. The "or negative definite" might seem a bit odd, but it is because involutions only correspond to bilinear forms up to multiplication by a scalar; in particular we can multiply the form by -1.

Hence the positivity of the Rosati involution says that if we think of the polarisation as being "like a bilinear form", then it is "positive or negative definite" (but -1 times a polarisation is not a polarisation, so we can treat polarisations as always being "positive definite").

The general proof that the Rosati involution is positive (valid over all fields, including those of positive characteristic) uses ├ętale cohomology to show that \operatorname{Tr}_{E/\mathbb{Q}}(\alpha \alpha^\dag) is equal to an intersection number involving ample divisors. I shall give a simpler proof over the complex numbers only, using the positive-definiteness of the Hermitian form associated with a polarisation.

Let V be the tangent space T_0 A(\mathbb{C}) and recall that a polarisation induces a positive-definite Hermitian form H on V. Any \alpha \in E induces a linear map \alpha_* \colon V \to V, and just as for the symplectic Riemann form, \dag is the adjoint involution with respect to H.

Lemma. If \alpha \in E - \{ 0 \}, then all eigenvalues of (\alpha \alpha^\dag)_* on V are nonnegative real numbers, and at least one is positive.

Proof. Let v \in V be an eigenvector of (\alpha \alpha^\dag)_* with eigenvalue \lambda. Then  H(\alpha^\dag_* v, \alpha^\dag_* v) = H((\alpha \alpha^\dag)_* v, v) = \lambda H(v, v). Because H is positive definite, H(\alpha^\dag_* v, \alpha^\dag_* v) is a nonnegative real number and H(v, v) is a positive real number, so \lambda is a nonnegative real number.

Now consider E acting on \operatorname{End}_\mathbb{C} V by left multiplication. The eigenvalues of \alpha \alpha^\dag on \operatorname{End}_\mathbb{C} V are simply its eigenvalues on V, repeated \dim V times.

Finally \beta \mapsto \beta_* embeds E \otimes_\mathbb{Q} \mathbb{C} as a subspace of \operatorname{End}_\mathbb{C} V which is stable under left multiplication by E. The eigenvalues of \alpha \alpha^\dag acting on this subspace must be a subset of its eigenvalues on \operatorname{End}_\mathbb{C} V, so are all nonnegative real numbers.

We note also that \alpha \alpha^\dag \neq 0. This is because \alpha^\dag \neq 0 so there is some v \in V such that \alpha^\dag v \neq 0. Then  H((\alpha \alpha^\dag)_* v, v) = H(\alpha^\dag_* v, \alpha^\dag_* v) \neq 0 so (\alpha \alpha^\dag)_* v \neq 0.

The trace of \alpha \alpha^\dag is the sum of its eigenvalues on E. We have shown that these are all nonnegative, and becaue \alpha \alpha^\dag \neq 0, at least one eigenvalue is nonzero. Hence the sum of the eigenvalues is positive.

Tags abelian-varieties, alg-geom, maths


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