Last time: to classify nondegenerate symmetric bilinear forms on vector spaces over or : find such that , scale it to get such that (), resp. (), then continue with . This gives a basis with
so
The skew-symmetric case
What about the skew-symmetric case? (Suppose .)
We can still find a “standard basis” for a finite-dimensional vector space with a nondegenerate skew-symmetric bilinear form (a.k.a. a symplectic form), but the process is slightly different since for all .
Instead: pick any nonzero ; since is nondegenerate, is nonzero, so there exists such that , and we can make it by scaling .
Now we find (if has then ), so
and we study the restriction of to the latter subspace (induction on dimension).
Proposition 18.1. Let be finite-dimensional over ,, and a nondegenerate skew-symmetric bilinear form . Then is even, and has a basis such that
i.e. the matrix of is block diagonal:
The group of linear transformations preserving is called the symplectic group.
Tensor products
Next: tensor products & multilinear algebra.
Goal: think of bilinear maps as linear maps from a new vector space .
For finite-dimensional vector spaces over , the tensor product is a vector space together with a bilinear map
Three definitions (from concrete to abstract; all are equivalent, i.e. give the same output up to natural isomorphism):
Definition 18.2 (Definition 1). Choose bases of and of . Then is the vector space with basis . The bilinear map is , extended by linearity.
Definition 18.3. Elements of the form
are called pure tensors; not every element of is of this form!
The rank of an element of is the minimal number of terms needed to express it as a linear combination of pure tensors.
This is concrete and makes it clear that , but the independence of the choice of basis isn’t obvious. To de-emphasize the basis:
Definition 18.4 (Definition 2). Start with a vector space with basis (this is insanely large: usually this basis is uncountable!), and quotient it by a subspace of relations among these elements: is the span of
Defining sets all these to zero, enforcing bilinearity of the map .
This shows independence of the basis, but involves an unpleasantly large construction. (At the end, if we have bases of and of , the relations in do show all elements of are linear combinations of the , but before one checks this, it’s not even obvious that .)
The least concrete, yet most mathematically satisfactory definition characterizes what does without spelling out how it’s actually constructed: namely, that linear maps from to another space, when evaluated on pure tensors , give maps from that are bilinear in and . (E.g. in Definition 2: is too big, and the quotient by enforces bilinearity.)
Definition 18.5 (Definition 3). The tensor product is the universal vector space through which all bilinear maps from factor, i.e. it is a vector space together with a bilinear map such that, given any vector space over and any bilinear map , there exists a unique linear map such that .
This tells us the key property of and implies uniqueness up to isomorphism (the universal property gives isomorphisms between any two candidate constructions of ), but existence ultimately comes from one of the previous constructions!
Check: Definition 1 satisfies the property: given bases and of and ,
by defining , and vice versa.
Basic properties
is a functor. This means: given linear maps and , we get a linear map , given on pure elements by , and this respects composition.
(natural isomorphism ).
.
More surprising but extremely useful:
Proof. The map
is bilinear, so by the universal property we get a linear map , which takes .
Pick bases of and of , and let be the dual basis of . Then is a basis of .
The above construction takes to
whose action on basis vectors is: maps to , all others to .
Thus is the matrix with a single nonzero entry, a in row , column . These form a basis of .
Since it maps a basis to a basis, is an isomorphism. ∎
Example 18.6. If has basis , has basis , and has basis , then the linear map with matrix is
This is in general a rank 2 tensor, except if : then we can write it as a pure tensor .
Proposition 18.7 (Fact). Tensor rank in is the same as rank in ! (Hence the name.)
(Rank 1 case: corresponds to , whose image is !)
This is easiest to see if we take a basis of in which is a basis of , and a basis of in which is a basis of , with for all . Then corresponds to
The isomorphism also implies:
. We can view this as:
.
This is actually the transpose construction . (Easiest to check on rank 1: , i.e. .)
Notes not reached in this lecture, covered at the start of {{LINK:L19}}
The trace, properly defined
We can now properly define the trace of a linear operator!
In “ordinary” linear algebra classes, one defines the trace of an matrix to be , the sum of the diagonal entries; then, noting that
we have , and so the trace of is defined to be the trace of in any basis.
We could also try to define the trace via eigenvalues and their multiplicities, over an algebraically closed field: in a basis where is triangular it is manifest that .
We can do better (conceptually), by using and the contraction linear map . Namely, there’s a natural bilinear pairing
and it determines , given on pure tensors by .
This is indeed equivalent to the usual definition: choosing a basis and the dual basis , the trace of the matrix with single entry 1 in position .
Multilinear maps and the tensor algebra
Definition 18.8. A map is multilinear if it is linear in each variable separately.
The tensor product can be defined as above, either using bases of , or as a quotient of a universal vector space by relations, or via the universal property for multilinear maps: there is a multilinear map
such that for every vector space and every multilinear , there exists a unique such that .
In fact nothing new is happening, because .
In the special case of
we have bilinear maps for all , which taken together define a multiplication on the tensor algebra
making it a noncommutative ring.