The Unapologetic Mathematician

Mathematics for the interested outsider

Multilinear Functionals

Okay, time for a diversion from all this calculus. Don’t worry, there’s tons more ahead.

We’re going to need some geometric concepts tied to linear algebra, and before we get into that we need to revisit an old topic: tensor powers and the subspaces of symmetric and antisymmetric tensors. Specifically, how do all of these interact with duals. Through these post we’ll be working with a vector space V over a field \mathbb{F}, which at times will be assumed to be finite-dimensional, but will not always be.

First, we remember that elements of the dual space V^*=\hom_\mathbb{F}(V,\mathbb{F}) are called “linear functionals”. These are \mathbb{F}-linear functions from the vector space V to the base field \mathbb{F}. Similarly, a “n-multilinear functional” is a function f that takes n vectors from V and gives back a field element in \mathbb{F} in a way that’s \mathbb{F}-linear in each variable. That is,

\displaystyle f(v_1,\dots,av_i+bw_i,\dots,v_n)=af(v_1,\dots,v_i,\dots,v_n)+bf(v_1,\dots,w_i,\dots,v_n)

for scalars a and b, and for any index i. Equivalently, by the defining universal property of tensor products, this is equivalent to a linear function f:V^{\otimes n}\rightarrow\mathbb{F} — a linear functional on V^{\otimes n}. That is, the space of n-multilinear functionals is the dual space \left(V^{\otimes n}\right)^*.

There’s a good way to come up with n-multilinear functionals. Just take n linear functionals and sew them together. That is, if we have an n-tuple of functionals (\lambda^1,\dots,\lambda^n)\in\left(V^*\right)^{\times n} we can define an n-multilinear functional by the formula

\displaystyle\left[m(\lambda^1,\dots,\lambda^n)\right](v_1\otimes\dots\otimes v_n)=\prod\limits_{i=1}^n\lambda^i(v_i)

We just feed the ith tensorand v_i into the ith functional \lambda^i and multiply all the resulting field elements together. Since field multiplication is multilinear, so is this operation. Then the universal property of tensor products tells us that this mapping from n-tuples of linear functionals to n-multilinear functionals is equivalent to a unique linear map from the nth tensor power \left(V^*\right)^{\otimes n}\rightarrow\left(V^{\otimes n}\right)^*. It’s also easy to show that this map has a trivial kernel.

This is not to say that dualization and tensor powers commute. Indeed, in general this map is a proper monomorphism. But it turns out that if V is finite-dimensional, then it’s actually an isomorphism. Just count the dimensions — if V has dimension d then each space has dimension d^n — and use the rank-nullity theorem to see that they must be isomorphic. That is, every n-multilinear functional is a linear combination of the ones we can construct from n-tuples of linear functionals.

Now we can specialize this result. We define a multilinear functional to be symmetric if its value is unchanged when we swap two of its inputs. Equivalently, it commutes with the symmetrizer. That is, it must kill everything that the symmetrizer kills, and so must really define a linear functional on the subspace of symmetric tensors. That is, the space of symmetric n-multilinear functionals is the dual space \left(S^nV\right)^*. We can construct such symmetric multilinear functionals by taking n linear functionals as before and symmetrizing them. This gives a monomorphism S^n\left(V^*\right)\rightarrow\left(S^nV\right)^*, which is an isomorphism if V is finite-dimensional.

Similarly, we define a multilinear functional to be asymmetric or “alternating” if its value changes sign when we swap two of its inputs. Then it commutes with the antisymmetrizer, must kill everything the antisymmetrizer kills, and descends to a linear functional on the subspace of antisymmetric tensors. As before, we can construct just such an antisymmetric n-multilinear functional by antisymmetrizing n linear functionals, and get a monomorphism A^n\left(V^*\right)\rightarrow\left(A^nV\right)^*. And yet again, this map is an isomorphism if V is finite-dimensional.

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October 22, 2009 - Posted by | Algebra, Linear Algebra


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  6. Hi, great site! I am still dizzy with all this duality. A couple of simple questions, please
    1)We have that (VxW)*~V*(x)W* (V,W finite dimensional vector spaces over the same field). Question: How do we evaluate a functional (w*x v*)(v,w)? Is it w*(w)+v*(v) ? I checked this operation is linear in pairs (v,w) , but this is no guarantee.

    2)Say G , G’ are Abelian groups, so that we can see them as Z-modules (Z=integers). Does a linear map L: G–>G’ also “induce” a map L*: G’* –>G* , as in the case of vector spaces? Is there a way of having an induced map seeing G,G’ as Abelian groups only, not as Z-modules (so that there is a reasonable definition of the dual space of an Abelian group)? .

    Comment by Larry | June 20, 2014 | Reply

  7. Good questions, Larry. As to the first one, it’s actually not addition but multiplication. If you have f\in V^* and g\in W^* then we evaluate (f\otimes g)(v, w) by f(v)*g(w). Notice how if W=V we’re in the situation in the post above for n=2: we have a 2-tuple (pair) of functionals (f, g)\in \left(V^*\right)^{\times 2}.

    As to the second, yes, an Abelian group is exactly the same thing as a \mathbb{Z}-module, so there’s no real way to see G as “just an Abelian group”. Duality only depends on the module structure of V — I addressed dual modules in an early post — so there’s definitely a dual module. The dual construction is still a contravariant functor, which means that yes, if you’ve got a function L: G\to H you get a dual function L^*: H^*\to G^*.

    To be explicit about it, remember that a functional \lambda\in H^* is a function \lambda: H\to\mathbb{Z}. You can clearly define the composite function \lambda\circ L: G\to\mathbb{Z}, which lets us define L^*(\lambda)=\lambda\circ L\in G^*.

    Comment by John Armstrong | June 21, 2014 | Reply

  8. Yes, sorry, re Abelian groups as Z-modules, I was thinking of the case where besides the group structure we also had a topological structure, so we could talk about a continuous dual, but I thought this may be too far from the topic you are covering in this post. Thanks for the answer.

    Comment by Larry | June 23, 2014 | Reply

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