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Added: are all types of mappings for vector spaces with "product" in their names always multilinear mappings between some vector spaces? Are there many counterexamples?

$F$ is a field. Any bilinear form on $F^n$ can be expressed as $$ B(\textbf{x},\textbf{y}) = \textbf{x}^\mathrm T A\textbf{y} = \sum_{i,j=1}^n a_{ij} x_i y_j $$ where $A$ is an n × n matrix.

I was wondering if $\textbf{x}^\mathrm T A\textbf{y}$ is a single product of some type between two vectors in $F^n$? What type is it?

More generally, is a multilinear form on $F^n$ a single product of some type on multiple vectors in $F^n$?

Is a multilinear form defined on the product space of a vector space a single product of some type on multiple vectors in the vector space?

Is a multilinear form defined from the product space of a vector space to another vector space a single product of some type on multiple vectors in the first vector space?

Thanks and regards!

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  • $\begingroup$ If you define "product" to include this, then yes! $\endgroup$ – Jonas Meyer Jan 16 '12 at 22:48
  • $\begingroup$ @JonasMeyer: Thanks! I was thinking maybe they call it tensor product or something defined in another way? Do they? $\endgroup$ – Tim Jan 16 '12 at 22:49
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Well, the overriding question here is what you mean by "product". Generally, a "product" on some structure $V$ (in this case, a vector space) is a binary operation of a sort; i.e. a mapping $$ V \times V \rightarrow V, $$ possibly satisfying some conditions (i.e. associativity, commutativity). As you have things set up, a bilinear form isn't going to be a "product" in this sense, because the codomain isn't right: a bilinear form eats two vectors and spits out an element of the base field, not another element of $V$, as would be the case for some sort of "product".

A very natural type of product operation that does arise on vector spaces is a Lie bracket, which generalizes things like the cross product, which really is a binary operation on your space which gives you a way to "multiply" vectors. An important feature of a Lie bracket is that it is neither commutative nor associative (but the situation isn't too bad, as it is anti-commutative and satisfies the Jacobi identity).

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  • $\begingroup$ Using "the" to modify "Lie bracket" seems a little misleading to me. On a given vector space one may define many Lie brackets. $\endgroup$ – Qiaochu Yuan Jan 17 '12 at 0:00
  • $\begingroup$ Thanks! I think tensor product is a kind of product does not give result in the same vector space as its operands. My question is: are all types of mappings with "product" in their names always multilinear mappings between vector spaces? $\endgroup$ – Tim Jan 17 '12 at 0:04
  • $\begingroup$ @QiaochuYuan. Good point. Amended. $\endgroup$ – NKS Jan 17 '12 at 0:06
  • $\begingroup$ @Tim: Well, the short answer is "no", but it really depends on what you mean by "all". Certainly once we stop doing linear algebra (and so don't even have a notion of "multilinearity") there will be products which show up that aren't of this type. But the tensor product, along with the symmetric and exterior products, are "products" in the sense that you speak of. $\endgroup$ – NKS Jan 17 '12 at 0:19
  • $\begingroup$ (continued) Incidentally, a good way to think about this type of "product" is that it is a way of turning multilinear maps into plain linear maps: if you have a bilinear map from a space V into W, you get a linear map from $V \otimes V$ into W. Adding the condition that this map be symmetric or anti-symmetric gives the symmetric and exterior product, respectively. And the bilinear forms that we started this discussion with can be thought of as elements in the dual of the tensor of $F^n$ with itself. $\endgroup$ – NKS Jan 17 '12 at 0:24
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Please correct me if I'm wrong, but aren't we just talking about contraction here between two vectors? Say $$\sum_{j=1}^{n}a_{ij}y_{j}=z_{i}$$ then $$\sum_{i,j=1}^{n}a_{ij}x_i y_j=\sum_{i=1}^{n}z_i x_i$$ which is an operation between two vectors. So I guess once again it depends on the sense of the word "product" for this case as stated before, but from what's been written here it looks like this maps to $F$, not $F^n$.

Does this help?

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  • $\begingroup$ Thanks! Could you define "contraction between two vectors"? $\endgroup$ – Tim Jan 17 '12 at 1:03
  • $\begingroup$ Sure, I've always thought of it as essentially the "dot product", but it is the multiplication between the same components of n-tuples, summed. Generally I see it in physics with the "summation convention" between vectors and co-vectors. Is this clear? $\endgroup$ – kηives Jan 17 '12 at 6:07

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