6
$\begingroup$

Is this convention? If we let $B$ be the space of bilinear maps, it seems to me that there is an isomorphism $i_1 : B \to V^* \otimes V^*$ and an isomorphism $i_2 : B \to V \otimes V$. So is it just convention to think about bilinear maps as the tensor product of the dual space, or are there reasons using $i_2$ doesn't make sense?

The spirit of this question is a possible duplicate of this question https://physics.stackexchange.com/questions/105347/is-it-foolish-to-distinguish-between-covariant-and-contravariant-vectors. But the answers there seem to be somewhat intertwined with concepts relating to general relativity. Is this a concept that only makes sense to understand in the context of differential geometry? If not, I am looking for answers that are more abstract.

$\endgroup$
10
  • $\begingroup$ It is in the light of the fact that a bilinear form is usually defined as a mapping from $V \times V$, i.e. it takes vectors (contravariant objects) as arguments. Since the bilinearity means linearity in each slot, you can understand each slot of a bilinear form to be a one-form (covariant object), i.e. element of $V^*$. Thus the identification with $V^* \otimes V^*$. $\endgroup$ Commented Feb 4, 2018 at 21:32
  • 4
    $\begingroup$ The isomorphism with $V \otimes V$ would vary depending on a choice of basis of $V$, whereas the isomorphism with $V^* \otimes V^*$, or with $(V \otimes V)^*$, wouldn't. $\endgroup$ Commented Feb 4, 2018 at 21:33
  • $\begingroup$ @DanielSchepler has a very good point, one of the isomorphisms is canonical whereas the other is not. $\endgroup$ Commented Feb 4, 2018 at 21:36
  • $\begingroup$ @RadekSuchánek What do you mean by "canonical"? $\endgroup$ Commented Feb 4, 2018 at 21:46
  • $\begingroup$ What @DanielSchelper said in his comment is the meaning of "canonical", the independence on the choice of basis in this context $\endgroup$ Commented Feb 4, 2018 at 23:04

2 Answers 2

6
$\begingroup$

By the universal property of the tensor product, every bilinear map $V \times V \to k$, where $k$ is the underlying field, corresponds uniquely to a linear map $V \otimes V \to k$. So almost by definition of the tensor product, the set of bilinear maps out of $V \times V$ is the space $Hom_k(V \otimes V, k) =: (V \otimes V)^*$.

Be careful though, it's true that if $V$ is finite dimensional we have isomorphisms $(V \otimes V)^* \cong V^* \otimes V ^* \cong V \otimes V$. However, when $V$ is infinite dimensional we don't have $V \cong V^*$, so that the identification of $(V \otimes V)^*$ with $V \otimes V$ is no longer true.

$\endgroup$
2
  • $\begingroup$ Could you explain the part about how when $V$ is infinite dimensional we don't have $V \cong V^*$? $\endgroup$ Commented Feb 4, 2018 at 21:56
  • 2
    $\begingroup$ @WilliamOliver In general, if $V$ is infinite dimensional then $dim(V^*) > dim(V)$. One can see this through the isomorphism $Hom_k(\bigoplus_{i \in I} V_i, k) \cong \prod_{i \in I} Hom_k(V_i, k)$ and maybe an application of the axiom of choice. This means that $(V \otimes V)^*$ (i.e. $BilinearMaps(V \times V, k)$) is in general bigger than $V \otimes V$. $\endgroup$
    – Exit path
    Commented Feb 4, 2018 at 21:59
4
$\begingroup$

Since $V^*$ is formed by the set of all linear maps from $V$ to the underlying field $\mathbb{F}$. It is natural to use two such objects tensored to form a bilinear map. It is just a constructive convenience, $\alpha \otimes \beta$ for $\alpha, \beta \in V^*$ forms a bilinear map on $V$ in the sense $\alpha \otimes \beta: V \times V \rightarrow \mathbb{F}$ is linear in both slots: in particular we define the $\otimes$ product of covectors $\alpha, \beta \in V^*$ by $$ (\alpha \otimes \beta)(x,y) = \alpha(x) \beta(y) $$ for all $x,y \in V$. It's easy to show bilinearity of $\alpha \otimes \beta$. Furthermore, we can build a basis for the space of bilinear forms on $V$ by the tensor product of a given dual basis to $V$. For basis $\beta = \{ e_1, \dots ,e_n \}$ we define $\beta^* = \{ e^1, \dots , e^n \}$ by $e^i(e_j) = \delta_{ij}$ extended linearly. Then, $e^i \otimes e^j$ for $1 \leq i,j \leq n = \text{dim}(V)$ forms a basis for the bilinear maps on $V$. In this sense the isomorphism of bilinear maps and $V^* \otimes V^*$ is natural. But, from an abstract perspective, the spaces $V \otimes V, V^* \otimes V$, $V \otimes V^*$ and $V^* \otimes V^*$ are all isomorphic. But, only $V^* \otimes V^*$ matches a bilinear form in the constructive manner I next describe.

To be explicit, given a bilinear form $b: V \times V \rightarrow \mathbb{F}$ we may construct it via: $$ b = \sum_{i,j} b(e_i,e_j) e^i \otimes e^j. $$ You're right, this is not something special to GR, it's just linear algebra.

For another example, consider $T: V \rightarrow V$ a linear transformation. We find natural identification with $V^* \otimes V$ since we can write $T = \sum_{i,j} e^j(T(e_i)) e^i \otimes e_j$ with the understanding $T(x) = \sum_{i,j} e^j(T(e_i)) e^i(x) \otimes e_j$. Here $e^j(x)$ picks off the $j$-th component of the vector $x$. Notice the matrix of $T$ is given by $e^j(T(e_i))$ whereas the matrix of the bilinear map $b(e_i,e_j)$ has a different structure which would be quite apparent under coordinate change. That's part of the trouble, isomorphism in the purely linear algebraic sense does not include the oft important nature of coordinate change. The matrix of a bilinear form and the matrix of a linear transformation on $V$ behave differently under coordinate change although the vector space of linear maps on $V$ and bilinear forms on $V$ are isomorphic.

$\endgroup$
2
  • $\begingroup$ So, in summary, it is technically convention from a purely abstract perspective, but it is far more intuitive and natural to make bilinear maps out of linear maps than out of the objects in their domains? $\endgroup$ Commented Feb 4, 2018 at 21:53
  • 1
    $\begingroup$ @WilliamOliver right, see the other answer for more how an algebraist likes to look at it. My perspective is more from a concrete formula building perspective, which is somewhat closer to the physicist's manner of thinking. Of course, the other answer is more mathematically elegant. $\endgroup$ Commented Feb 4, 2018 at 21:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .