S. D. V. Stephens

notes / Studies in Algebra and Group Theory / chapter 4: Categories, Bilinear Forms and Inner Products

14 Functors, Bilinear Forms, and Inner Products

· Denis Auroux

Original Professor Auroux’s handwritten notes (PDF), behind a password for now:

Interlude: the language of categories

(Then we’ll return to (bi)linear algebra.)

Definition 14.1. A category is a collection of objects, plus for each pair of objects a collection of morphisms , and an operation called composition of morphisms, such that

  1. every object has an identity morphism such that for all , ;
  2. composition is associative: .

Example 14.2.

  1. The category of sets, = all maps .
  2. : finite-dimensional vector spaces over , linear maps.
  3. Groups, group homomorphisms.
  4. Topological spaces, continuous maps.

Definition 14.3. is an isomorphism if there exists (the inverse isomorphism) such that and .

Check:

  • the inverse of , if it exists, is unique;
  • is an isomorphism; iso iso; isos iso.

So the automorphisms of , , form a group.

  • Isomorphic objects have isomorphic automorphism groups: an isomorphism determines an isomorphism of groups , .

Example 14.4.

  1. In Sets, a finite set with elements .
  2. an -dimensional vector space over , the invertible matrices.

Category theory language can be used to formalize the common features of sums, products, quotients of objects in different categories. E.g. a product object is characterized by: there exist morphisms and (“projections”) such that for every object , for all and , there exists a unique such that and . (In the usual categories, and , .)

We won’t study this further.

Functors

Definition 14.5. Let be categories. A (covariant) functor is an assignment

  • to each object in , of an object in ;
  • to each morphism , of a morphism ,

such that

  1. ;
  2. .

Example 14.6.

  1. The forgetful functor taking a group, a topological space, … to the underlying set.
  2. On vector spaces: given a vector space , . If is linear, then the induced map is This gives a functor (denoted ).
  3. Complexification, : on objects, ; on morphisms, (seen last time).

A contravariant functor is the same except the direction of morphisms is reversed:

Example 14.7. On , (the dual) (see HW 5).

There’s one more layer to this, if you love category theory: given 2 functors , a natural transformation from to is the data, for all , of a morphism , such that for all and all , commutes in .

Example 14.8. On , (double dual) is a (covariant) functor. We’ve said there is a “natural” map (an isomorphism if ). The precise meaning is: is part of a natural transformation of functors , from the identity functor to the double dual functor (see HW 5).

Bilinear forms

Definition 14.9. A bilinear form on a vector space over a field is a map that is linear in each variable: for all and all ,

This is not a linear map ().

Definition 14.10. We say is symmetric if for all , and skew-symmetric if .

Example 14.11.

  • The usual dot product on , , is symmetric.
  • , is skew-symmetric.

Given a bilinear map , we get a linear map by defining (it maps to ). Conversely, determines the bilinear form . This defines a bijection

Definition 14.12. The rank of is the rank of (). If is an isomorphism, we say is nondegenerate.

For a given vector space , is a vector space over . What is its dimension? If we choose a basis for , it is enough to specify for all in order to determine : by bilinearity, The values of can be chosen freely; e.g. a basis of is given by , So (consistent with ! The bijection is an isomorphism of vector spaces!).

Given a basis of , is represented by an matrix : where is the matrix of , . So, in terms of column vectors,

Remark (Skip this remark if your head hurts). The isomorphism , , is natural, in the sense that: we have contravariant functors and (on morphisms, gives and the isomorphisms define a natural transformation between them.

Notes not reached in this lecture, covered at the start of {{LINK:L15}}

Definition 14.13. If is a subspace of a vector space equipped with a bilinear form , we define its orthogonal complement This is a vector subspace.

Warning: this is most intuitive if is symmetric or skew. Otherwise we have to worry about “left-orthogonal” vs. “right-orthogonal” to .

Lemma 14.14. If is nondegenerate then (else ).

Proof. where this map is the composition of and the restriction , . By the rank theorem, . If is nondegenerate then is an isomorphism and is surjective, so ; in general . ∎

Example 14.15.

  • with the standard dot product : then , the “usual” orthogonal complement, because (see below) and .
  • But for , if is the 1-dimensional subspace spanned by any nonzero vector , then !! (Because .)

Inner product spaces

Definition 14.16. An inner product space is a vector space over together with a symmetric, definite positive bilinear form .

Symmetric: . Definite positive: for all , and iff .

This definition only makes sense over an ordered field, so that “” makes sense. In practice this means . We can’t do this over (we’ll see a workaround: Hermitian forms).

  • Let , , be the linear map corresponding to . Then definite positive is injective (since for all , ! ) (assuming ) is an iso , i.e. is nondegenerate. (The converse is false: nondegenerate positive.)

Proposition 14.17. a finite-dimensional inner product space, a subspace .

Proof.

  • We’ve seen: is nondegenerate, so .
  • Since is positive definite, . So . Since the dimensions add up to , this implies .

∎

Definition 14.18.

  • The norm of a vector is .
  • are orthogonal if .

Observe: .

  • If and are orthogonal then (Pythagorean theorem).
  • In general, by analogy with the law of triangles, we define the angle between 2 vectors, This only makes sense if ?

Theorem 14.19 (Cauchy–Schwarz inequality). For all , .

Proof. The inequality is unaffected by scaling, so we can assume . Decompose along where . Explicitly, , with and orthogonal to . Then , so This is the desired inequality for . ∎