Let $E_1$ and $E_2$ be two finite dimensional vector spaces over $\mathbb{R}$. Then is it true that the tensor product $E_1 \otimes E_2$, is defined as the set of all bilinear maps from $E_2 \times E_1$ to $\mathbb{R}$?

I ask this question since I know that $ V \otimes V^* $ is the space of all bilinear forms on $V^* \times V$, if $V$ is a finite dimensional vector space.


1 Answer 1


Not quite; the right way to go is not to "swap" the vector spaces $E_1,E_2$, but to take their dual. So the tensor product $E_1\otimes E_2$ should correspond to the set of all bilinear maps $\{E_1^*\times E_2^*\xrightarrow{\sim\ }\Bbb R\}$.

In general, if $E_1,\dotsc,E_n$ are finite-dimensional and $F$ is any vector space, $$E_1\otimes \cdots\otimes E_n \otimes F \simeq \{E_1^*\times \cdots\times E_n^*\xrightarrow{\sim\ } F\},$$ where $\simeq$ may mean equality or (canonical) isomorphism depending on your definitions. Additionally, note that:

  • The bidual $E^{**}=(E^*)^*$ of a finite-dimensional vector space $E$ is canonically isomorphic to $E$ itself (this is violently false for non-finite-dimensional vector spaces);
  • If $E$ is any real vector space, as is your case, then it is sensible to assume you are considering tensor products over the real field, and in that case we have the canonical isomorphisms $E\otimes \Bbb R \simeq E \simeq \Bbb R\otimes E$;
  • The tensor product is always associative, meaning that there are canonical isomorphisms $(E_1\otimes E_2)\otimes E_3\simeq E_1\otimes E_2\otimes E_3 \simeq E_1\otimes (E_2\otimes E_3)$ whatever the dimension of the vector spaces involved. (Basically, we can drop parentheses.)

The combination of all these facts leads to $$E\otimes E^* \simeq (E\otimes E^*)\otimes \Bbb R\simeq E\otimes E^*\otimes \Bbb R\simeq \{E^*\times (E^*)^* \xrightarrow{\sim\ }\Bbb R\}\simeq \{E^*\times E \xrightarrow{\sim\ }\Bbb R\},$$ which is the statement you recalled.

  • $\begingroup$ Ah, so we are identifying the double dual with the vector space itself! (possible in finite dimensions). Also, one can drop the field $F$ in the last tensor product entry in $E_1\otimes \cdots\otimes E_n \otimes F \simeq \{E_1^*\times \cdots\times E_n^*\overset{\sim}{\to} F\}$, if it is clear from the context, right? So $E\otimes E^*$ means the same as $E\otimes E^* \otimes F$, right? $\endgroup$ Commented May 9, 2023 at 19:54
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    $\begingroup$ $F$ is a unit object for $\otimes_F$, meaning that, for any $F$-module $V$, $V \otimes_F F \cong F \otimes_F V \cong V$, naturally in $V$ $\endgroup$ Commented May 9, 2023 at 20:00
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    $\begingroup$ What giobrach meant is that it holds for any vector space $F$, not just restricted to $F$ being 1-dimensional. $\endgroup$ Commented May 9, 2023 at 20:03
  • $\begingroup$ @Category_Theorist Thank you for the response. $\endgroup$ Commented May 9, 2023 at 20:14
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    $\begingroup$ @ApoorvPotnis I have updated my answer with more details. $\endgroup$
    – giobrach
    Commented May 10, 2023 at 13:29

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