When considering vector spaces $A$ and $B$, what is the difference between $A \times B,$ $A \otimes B$ and $A \wedge B?$

Question

I have looked at this resource http://hitoshi.berkeley.edu/221a/tensorproduct.pdf to instinctively differentiate between the tensor product and the direct sum of two vector spaces.

I am currently working on Lie Algebras. Some resources tell me that the Lie Bracket $[\cdot, \cdot]: \mathfrak{g} \times \mathfrak{g} \to \mathfrak{g},$ whereas some tell me that $[\cdot, \cdot]:\mathfrak{g} \otimes \mathfrak{g} \to \mathfrak{g}.$ Are these definitions in fact equivalent? If so why? What does the tensor product add information wise that the standard product does not?

Finally, is there any relation between the tensor product and the wedge product?

Answer 1

The Lie bracket is an alternating bilinear map. This means you can describe it in three different ways:

• As a function $\mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$ (which is alternating and bilinear),
• As a linear map $\mathfrak{g} \otimes \mathfrak{g} \to \mathfrak{g}$ (which is alternating), or
• As a linear map $\Lambda^2 \mathfrak{g} \to \mathfrak{g}$.

(Here I'm ignoring the Jacobi identity.) In the first piece of notation, $\mathfrak{g} \times \mathfrak{g}$ denotes a set, not a vector space. One reason to introduce the tensor product is to have the ability to talk about bilinear maps in a way which stays entirely "inside" vector spaces, without mentioning sets.

There is no such thing as the wedge product of two vector spaces. There is such a thing as the exterior powers $\Lambda^k V$ of a vector space. I guess some people denote $\Lambda^2 V$ by $V \wedge V$, but in my opinion this is a very bad idea, as it suggests (incorrectly) that there's such a thing as $V \wedge W$.