2
$\begingroup$

In the classic Finite-Dimensional Vector Spaces by P. Halmos he defines the Tensor product as

The tensor product $U \otimes V$ of two finite-dimensional vector spaces $U$ and $V$ (over the same field) is the dual of the vector space of all bilinear forms on $U \oplus V$ [remark, by this he means all bilinear maps $\omega : U \times V \to \mathbb F$]. For each pair $x$ and $y$, with $x$ in $U$ and $y$ in $V$, the tensor product $z = x \otimes y$ of $x$ and $y$ is the element of $U \oplus V$ defined by $z(w) = w(x,y)$ for every bilinear form $w$.

Another construction, for example here or on Wikipedia, the tensor product is defined as the set of all formal sums on $U \times V$ modulo certain (''bilinear'') identities. This is equivalent to the set of all formal sums on $\{ u_i \}_{i\in I} \times \{ v_j \}_{j \in J}$ for bases $u_i$ and $v_j$ of $U$ and $V$.

According to my first reference, the set of all multilinear maps $V_1 \times \ldots \times V_k \to \mathbb F$ could be identified with $V_1^* \otimes \ldots \otimes V_k^*$, this seems always possible (not requiring finite dimensions), so in particular the set of all bilinear maps $U \times V \to \mathbb F$ could be identified with $V^* \otimes U^*$, so in Halmos definition the tensor product equals $(V^* \otimes U^*)^*$, where the tensor product is defined as the set of formals sums as above.

But to make sense, this definition must equal the definition by formal sums, so we must have $(V^* \otimes U^*)^* \cong V \otimes U$ to bring Halmos definition and the definition by formal sums in harmony.

But I guess here seems to be some sort of reflexivity involved, so he might used the finite-dimensionality of the vector spaces, or could Halmos definiton further extended, does I got it right?

Still another construction I found is by the following universal property (see also Wikipedia), i.e. the tensor product of two vector spaces $U$ and $V$ is such that for each bilinear map $U \times V \to W$ there exists exactly one linear map $U \otimes V \to W$, and this correspondence is linear, i.e. we have $$ \mbox{Bil}(U, V; W) \cong \mbox{Hom}(U\otimes V, W). $$ In our case $W = \mathbb F$ the ground field, then $$ \mbox{Bil}(U, V; \mathbb F) \cong (U\otimes V)^* $$ But by Halmos definition the tensor product equals $\mbox{Bil}(U, V; \mathbb F)^*$, so again this seems require some sort of reflexivity, i.e. $((U \otimes V)^*)^* \cong (U\otimes V)$.

But here the property to bring the definitions together reads still a little bit different, to make them all the same we need $(U^* \otimes V^*) \cong (U\otimes V)^*$, but this holds in general vector spaces, as $(U^* \otimes V^*) \cong \mbox{Bil}(U, V; F) \cong (U \otimes V)^*$.

So my questions again: 1) does I got everything right?, 2) am I right that Halmos definition is limited, and how far could we go with it, is reflexivity the right property to demand?

$\endgroup$
2
$\begingroup$

Yes, you've got it right. Remember the title of Halmos's book is Finite dimensional vector spaces! My guess is he probably thought the definition as the dual of bilinear forms was less abstract that either the quotient or the universal property definitions and used that one.

As you've realized Halmos's definition only agrees with the others when the spaces are reflexive, and since here we are talking about the algebraic dual (i.e., we don't have some norm and take dual to consist of bounded linear functionals, rather we are talking about all linear functionals), reflexive spaces are exactly the finite dimensional ones.

$\endgroup$
  • $\begingroup$ Thanks for the answer, so now I am more sure about the concepts and their relations. But one question left, when he has no problem using reflexivity, why not defining it as the space of all bilinear forms on $U\times V$ from the start, thus avoiding the extra dual, keeping it still simpler? $\endgroup$ – StefanH Oct 21 '14 at 12:37
  • $\begingroup$ Taking just bilinear forms gives a functor of the wrong variance, @Stefan: given linear maps $U \to U'$ and $V \to V'$, what you naturally get is a linear map $\mathrm{Bil}(U',V';\mathbb{F}) \to \mathrm{Bil}(U,V; \mathbb{F})$. This is contravariant, but you want the tensor product to be a covariant functor. $\endgroup$ – Omar Antolín-Camarena Oct 21 '14 at 13:10
  • $\begingroup$ Now looking through my literatur I found that for example Spivak in Calculus on Manifolds defines tensors directly as multilinear maps, thus defining the space of all $2$-tensors as the set of all bilinear forms on $V\times V$, and proceeding to show that this set is exactly $V^* \oplus V^*$, where $V^* \oplus V^* := \{ \varphi(u)\cdot \psi(v) : \varphi, \psi \in V^* \} \subseteq \mathcal I^2(V) := \{ \varphi : V\times V \to \mathbb K \mbox{ is bilinear } \}$. So here he uses also reflexivity in some sort, but defines $2$-tensors as the space of all bilinear forms... $\endgroup$ – StefanH Oct 21 '14 at 13:18
  • $\begingroup$ ... the approach (from Spivak, and which I mentioned in my first comment) you said is inadequate. Unluckily I am not that used to co-/contravariance, still learning, but in view of Spivaks approach in the finite-dimensional setting is this co-/contravariance thing not a big issue, i.e. it collapses? $\endgroup$ – StefanH Oct 21 '14 at 13:20
  • 1
    $\begingroup$ I think something different is the issue. In differential geoemtry people makes use of both maps $V \otimes V \to \mathbb{R}$ and elements of $V \otimes V$. The first are called 2-tensors (there are also $k$-tensors for all $k$) and the second are called bivectors (and for larger $k$, polyvectors). These tow spaces, $(V \otimes V)^\ast \cong V^\ast \otimes V^\ast$ (for finite dimensional $V$, which is the case that arises for finite dimensional manifolds), and $V \otimes V$ depend on $V$ with different variance. (to be cont'd) $\endgroup$ – Omar Antolín-Camarena Oct 21 '14 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.