The space of bilinear forms contains the symmetric and the alternating forms. This section carries out the analogous constructions for all tensor powers. Throughout, is a finite-dimensional vector space over , , and .
9.10.1 The Symmetric Group Acting on Tensors
For , the map , , is multilinear, so it induces a linear map , written : The factor in position moves to position . This is an action of on by linear maps (Definition 3.2.1). Indeed, for , writing , which is because . Both sides are linear, so for all , and clearly . (The inverses make this a left action for our convention that applies first.)
Definition 9.10.1 (Symmetric and alternating multilinear maps). A multilinear map (where ) is symmetric if and alternating if whenever for some . For , the alternating multilinear maps are the alternating forms of Definition 5.4.1, and for and the symmetric and alternating ones are the symmetric and alternating bilinear forms of Definition 9.1.2.
Lemma 9.10.2 (Alternating maps change sign). Let be multilinear and alternating.
- Exchanging two of the arguments of changes the sign of its value.
- for every .
- whenever are linearly dependent.
Conversely, if , a multilinear map with property (i) is alternating.
Proof. For alternating multilinear forms, that is, for , these are parts (a)–(c) of Lemma 5.4.2. Their proofs use only that the map is multilinear and alternating, never that its values are scalars, so they apply word for word to . For instance, (i) comes from expanding where the two terms with a repeated argument have been dropped. For the converse, if exchanging two arguments changes the sign, then a value with the same vector in two positions equals its own negative, so twice it is , and it is when in . ∎
9.10.2 Symmetric and Exterior Powers as Quotients
Let be the subspace spanned by the tensors for and , and let be the subspace spanned by the pure tensors with for some . For both are zero.
Definition 9.10.3 (Symmetric power). The th symmetric power of is the quotient space The class of is written .
Definition 9.10.4 (Exterior power). The th exterior power of is the quotient space The class of is written and called a wedge product.
In the factors of a product may be permuted freely, since lies in . In a wedge product with a repeated factor is zero. Since we get ; the same computation in any two positions shows that exchanging two factors of a wedge product changes its sign. For small , and .
Proposition 9.10.5 (Universal properties).
- The map , , is multilinear and symmetric. For every symmetric multilinear map there is a unique linear map with .
- The map , , is multilinear and alternating. For every alternating multilinear map there is a unique linear map with .
Consequently is isomorphic to the space of symmetric multilinear maps , and to the space of alternating multilinear maps .
Proof. The two maps are the composites of with the quotient maps, so they are multilinear. The first is symmetric because has the same class modulo as ; the second is alternating by the definition of .
(i) Given , let be the linear map with (Theorem 9.9.2). For each , the linear map vanishes on pure tensors because is symmetric, so it vanishes identically. Hence , and factors through the quotient .
(ii) If is alternating, vanishes on the spanning elements of , so it factors through .
In both cases is unique because the classes of pure tensors span the quotient. The last statement follows as for : is a linear bijection, with inverse given by composing with the map or . ∎
Proposition 9.10.6 (Induced maps). Let be a linear map between finite-dimensional spaces. There are unique linear maps and with They satisfy and , and likewise for .
Proof. The map is multilinear, and alternating because implies . Proposition 9.10.5 gives . Both sides of each identity are linear and agree on wedge products, which span. The case of is the same. ∎
These induced maps are what make it possible to form the symmetric and exterior powers of a representation, a notion defined in Chapter 13 (Definition 13.1.1).
9.10.3 Bases and Dimensions
Fix a basis of . For a -tuple of indices in , write by Theorem 9.9.2, these tensors form a basis of . For we have , where is with its entries rearranged.
Proposition 9.10.7 (Basis of ). For with integers and , let be the product of copies of , copies of , and so on. These monomials form a basis of , and
Proof. Spanning. The classes of the span . The class of is ; since its factors may be permuted, it equals , where is the number of entries of equal to .
Independence. Let be a vector space with a basis indexed by the exponent vectors of total , and let be the linear map with . Rearranging the entries of does not change , so for all and . Since the span and is linear, vanishes on , so it induces a linear map with . The are linearly independent, hence so are the .
Counting. Exponent vectors correspond to arrangements of stars and bars in a row ( is the number of stars between the st and the th bar), and there are such arrangements. ∎
Proposition 9.10.8 (Basis of ). For an increasing tuple , write . These elements form a basis of , so In particular for .
Proof. Let be the quotient map. Lemma 9.10.2(ii), applied to the alternating map , gives and ; by linearity, for all .
Spanning. The classes span . If has a repeated entry, . If the entries of are distinct, let be the increasing tuple with the same entries; there is a unique with , that is, , and then . Hence the span.
Independence. Let be a vector space with a basis indexed by the increasing -tuples, and let be the linear map with if has a repeated entry and otherwise (with and as above). We show that vanishes on . Let with for some , let , and write . Expanding, and because (the tuple is with its th and th entries exchanged). The terms with have . The remaining tuples come in pairs with , and : either both tuples have a repeated entry, or , so that and . So the contributions of each pair cancel, and . Thus , and induces a linear map with . The are independent, hence so are the .
If there are no increasing -tuples, so . ∎
Corollary 9.10.9 (The top exterior power). If , then is one-dimensional, and for every basis of the element is a basis of .
The line is called the determinant line of ; Theorem 9.11.3 below explains the name.
Note. Compare the three dimension formulas, for :
- , the number of ordered -tuples of indices;
- , the number of multisets of size from indices;
- , the number of subsets of size of indices.
Tensor powers remember the order of the factors; symmetric powers allow repeated factors and forget their order; exterior powers forbid repeated factors and remember their order only up to sign.
Proposition 9.10.10 (Wedge products detect linear independence). Vectors are linearly independent if and only if in .
Proof. If the are dependent, by Lemma 9.10.2(iii), applied to the alternating map . If they are independent, extend them to a basis of . Then is the basis element , , of Proposition 9.10.8 for this basis, so it is nonzero. ∎
9.10.4 Symmetric and Alternating Tensors
Many books define and as subspaces of . When is invertible in , the subspaces are isomorphic to our quotients.
Definition 9.10.11 (Symmetric and alternating tensors). A tensor is symmetric if for all , and alternating if for all . They form subspaces and of .
Proposition 9.10.12 (Tensor models when is invertible). Suppose in (for instance, ). Define linear maps by Then is a projection onto with kernel , and is a projection onto with kernel . Consequently and the quotient maps restrict to isomorphisms and , whose inverses are
Proof. We treat ; the argument for is the same with the signs omitted. For , substituting and using , so the image of lies in . If , each term equals , so . Hence is a projection onto , and .
Let be the quotient map. As in the proof of Proposition 9.10.8, , so ; that is, for every . If , this says . Conversely, let with for some , and let . Then , and substituting gives , so . As , divides , so in and . Hence . It follows that restricts to an isomorphism , with inverse . For this inverse gives , which is the stated formula after replacing by .
For , the identity shows , and shows . ∎
Note. Under these isomorphisms corresponds to the alternating tensor . Some authors identify with , without the factor ; this rescales the identification by . When in , the subspace models break down: Exercise 9.11.14(1) shows that over a field of characteristic , is three-dimensional, while is one-dimensional. This is why we define and as quotients.
Corollary 9.10.13 (Symmetric and alternating parts of ). Suppose , and let be the swap, . Then where acts as on and as on . Under the isomorphism of Proposition 9.8.13(ii), corresponds to the symmetric bilinear forms and to the skew-symmetric ones. Hence is the direct sum of the symmetric and the skew-symmetric forms, and
Proof. The swap is the action of the transposition in , so and . Since , every is the sum of an element of and an element of ; and if lies in both, then , so . This is also the case of Proposition 9.10.12, which gives the isomorphisms with and . Under Proposition 9.8.13(ii), corresponds to and to ; by linearity, corresponds to with , whose -eigenspaces are the symmetric and the skew-symmetric forms. Finally by Proposition 9.8.13(i), and we apply the first statement to . ∎
9.10.5 The Symmetric Algebra and Polynomials
The product of the tensor algebra descends to symmetric powers. For , the map given by is multilinear and symmetric in the for fixed , so by Proposition 9.10.5 it induces, for each fixed , a linear map . For a fixed , the image of is multilinear and symmetric in : this holds when , and such elements span . So it induces a linear map . The result is a bilinear product with
Definition 9.10.14 (Symmetric algebra). The symmetric algebra of is with the product that extends the products bilinearly.
The polynomial ring in variables is the vector space with basis the monomials , , with the multiplication extended bilinearly; for it is the ring . The homogeneous polynomials of degree are the linear combinations of the with .
Theorem 9.10.15 (The symmetric algebra is a polynomial ring). is a commutative ring with identity , and its multiplication is -bilinear. If is a basis of , the linear map is an isomorphism of rings. In particular, is isomorphic to the space of homogeneous polynomials of degree in variables.
Proof. The product is bilinear by construction, so it distributes over addition. Associativity can be checked on products of vectors, where both sides are ; commutativity too, since differs from only by a permutation of the factors. The element is an identity. By Proposition 9.10.7, the monomials (of all degrees) form a basis of , so the map is a linear isomorphism. It is multiplicative on basis elements, because (rearrange the factors) and ; both products are bilinear, so the map is multiplicative. ∎
Note. is the coordinate-free version of the polynomial ring: it depends only on , and a choice of basis identifies it with .
The most natural instance is . If is a basis of and , then is the th coordinate of , and a product of the should be a polynomial function of the coordinates. A function is a homogeneous polynomial function of degree if for a homogeneous polynomial of degree ; a linear change of coordinates turns homogeneous polynomials of degree into homogeneous polynomials of degree , so this does not depend on the basis.
Proposition 9.10.16 ( and polynomial functions). Let be a basis of and .
- There is a unique linear map from to the space of functions with .
- sends the monomial to the function . So the image of is the space of homogeneous polynomial functions of degree on .
- If is infinite, is injective, and so is the map it induces on . Thus is identified with the homogeneous polynomial functions of degree on , and with the ring of polynomial functions on , the variable being the th coordinate function.
Proof. (i) The map is multilinear and symmetric, with values in the vector space of functions ; apply Proposition 9.10.5. (ii) follows from (i).
(iii) Extend to degree by degree. By Theorem 9.10.15, an element of is a polynomial , and sends it to the function . So it suffices to show that a polynomial that vanishes at every point of is zero. We use induction on . For , a nonzero polynomial has at most roots (Corollary 4.3.15), while is infinite. For , write . For each point , the one-variable polynomial vanishes on all of , so every . Thus each vanishes on , and by induction. ∎
Note. When is finite, need not be injective. Over , the monomials and in define the same function, because for every . So over finite fields consists of polynomials as algebraic expressions, and these carry more information than the functions they define.
Note. When in , Propositions 9.10.12 and 9.9.6 identify with the symmetric tensors in , that is, with the symmetric -linear forms on . Following through these identifications gives the form and . So the polynomial function attached to a symmetric multilinear form is . For this is the passage from a symmetric bilinear form to its quadratic form (Section 9.6), and polarization (Proposition 9.6.2) recovers the form from the function.
9.10.6 The Exterior Algebra
Exactly as for symmetric powers, the map is multilinear and alternating in the for fixed , and in the for fixed . Applying Proposition 9.10.5 once in each variable gives a bilinear product with
Definition 9.10.17 (Exterior algebra). The exterior algebra (or Grassmann algebra) of is with the product that extends the products bilinearly. (Since for , the sum stops at .)
Proposition 9.10.18 (Properties of the exterior algebra). Let .
- is a ring with identity , its multiplication is -bilinear, and .
- The product is graded-commutative: for and ,
Proof. (i) Distributivity holds because the product is bilinear, and associativity and the identity can be checked on wedge products of vectors, where the product is concatenation. By Proposition 9.10.8, .
(ii) Both sides are bilinear in , so it suffices to take and . Starting from , move to the right past the factors , then , and so on. This reaches after exchanges of adjacent factors, each of which changes the sign (Lemma 9.10.2(i)). ∎
9.10.7 Operators on Tensor, Symmetric, and Exterior Powers
For a diagonalizable operator, the induced operators on tensor, symmetric and exterior powers are easy to describe; this is what Chapter 13 uses to compute the characters of and (Definition 13.4.1).
Proposition 9.10.19 (Eigenvalues on tensor, symmetric, and exterior powers). Let , and suppose is a basis of with . With the notation of Propositions 9.10.7 and 9.10.8:
- ;
- ;
- .
In particular , and are diagonalizable, and If , then
Proof. By the definitions of the induced maps, , and (i) and (ii) are the same computation. So each of the three bases consists of eigenvectors, and each trace is the sum of the eigenvalues. The monomials correspond to the nondecreasing tuples (list each index exactly times), which gives the formula for ; the sum over all -tuples of is . For , and , since . From we get , and . ∎