The construction of $V^* \otimes V^*$ involves creating formal symbols and then adding in relations such as bilinearity by quotienting out. A bilinear form $V\times V\to F$ can be thought of as a matrix by considering $y^T Mx$ and ignoring $y^T$ and $x$. Two linear forms $L: V \to F$ and $M: V \to F$ have a notion of a tensor product $(L \otimes M)(\alpha, \beta) = L(\alpha)M(\beta)$.

Is there suppose to be a correspondence between the tensors of $V^* \otimes V^*$ and the bilinear forms $V\times V\to F$. In this way the use of the term "tensor product" in the formal construction of the tensor product $V^* \otimes V^*$ is reconciled with the concrete notion of tensor products of forms.

But I can't reconcile the formal construction of the exterior product (or wedge product) with the concrete notion of exterior products for multilinear forms. On the one hand a book I am reading refers to the exterior product of an alternating $r$-linear form $L$ and an alternating $s$-linear form $M$using the tensor product

$$L \wedge M = \frac{1}{r!s!} \pi_{r+s}(L \otimes M)$$

where $\pi$ is defined generally as $\pi_rL=\sum _\sigma (\text{sgn } \sigma)L_{\sigma}$.

But on the other hand in the formal construction notion isn't the exterior product suppose to be the same as the tensor product but with extra relations created by quotienting?

How am I suppose to understand the relationship between the exterior product of two forms and the exterior product as taking the tensors and adding in new relations?


1 Answer 1


Antisymmetrization is an alternating function on tensors, and thus factors through the wedge product. The map

$$ a \wedge b \mapsto \frac{1}{2} \left(a \otimes b - b \otimes a\right) $$

has a one-sided inverse

$$ a \otimes b \mapsto a \wedge b $$

and thus the space of wedges is isomorphic to the space of anti-symmetric tensors. Some will take advantage of this and identify wedges with anti-symmetric tensors to avoid introducing more abstract constructions.

This topic becomes a lot more complicated in a setting where $2$ isn't invertible.

  • $\begingroup$ "This topic becomes a lot more complicated in a setting where 2 isn't invertible." Is that why a lot of these topics are almost always qualified with some statement about the characteristic of the base field? $\endgroup$
    – Muphrid
    Commented Jul 6, 2014 at 3:53
  • $\begingroup$ Yeah. Positive characteristic actually messes tons of stuff up. It also introduces neat new things as well. $\endgroup$
    – user14972
    Commented Jul 6, 2014 at 12:01

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