TITLE: Tensor product in multilinear algebra
QUESTION [5 upvotes]: In the book by Halmos ($FDVS$) the tensor product of two vector spaces U and V is defined as the dual of the vector space of all the bilinear forms on the direct sum of U and V. Is there a generalised form of this for the direct sums of more than two vector spaces? Is there a relation between the space of all multilinear forms on the direct sum of $V_1$,$V_2$,$V_3$,...,$V_k$ with their tensor product.
Please explain without invoking other algebraic objects such as modules,rings etc and by using the concepts regarding vector spaces only (as the book assumes no such background either, it is unlikely that any reader of that book will benefit from such an exposition). Everywhere I searched, I found the explanation in terms of those concepts only and being unfamiliar to those I couldn't get them at all.
Thanks in advance.

REPLY [3 votes]: (As in my comment at the top...) Sincerely and seriously, apart from whatever short-term adaptations may be necessary or desirable... As @RyanReich's answer and comments suggest, an "elementary" "definition" of tensor product will leave a person mystified, for one thing. For another, it's completely unclear why we need such a thing. (Be thankful you aren't studying the late-nineteenth-century version of linear and multi-linear algebra, with "upper and lower multi-indices", thus spawning "covectors" $v^i$, which somehow were different from "vectors" $v_i$. "Different"? In what way? Well, supposedly, in the "rules"... Nevermind.)
The long-term advantageous viewpoint, which is much better for far more reasons than just understanding tensor products per se, has two parts. The first is about the transition from "definition" in the sense of "construction" (usually in set-theoretic terms), to "characterization" in the sense of how the thing is intended/required to interact with other things. Yes, this inverts the lower-level textbook-y approach of the last several decades, in which we (somewhat disingenuously!) give a "definition" at the very beginning ... not admitting that people went through hosts of examples and many trial definition before hitting on the best abstraction ... and then "proving" the (desired/required) properties in an exaggeratedly formal (but stylish) style. In particular, for tensor products, while one has been conditioned to expect "a construction", in truth that's not what we truly want: we want to know what the point is, what's going to be accomplished. But a certain overly-formal textbook version of mathematics has scant way to say this, except for... (drum-roll)...
"Category theory". Informal/naive category theory is a wonderful thing, in the same way that informal/naive set theory is a wonderful thing, as it gives a language and family of concepts that enrich us, that broaden the possibilities, show commonalities, etc. One must be cautious about the more "formal/axiomatic" versions, however, at least that they do tend to be "X for its own sake" or "axiomatizations of tangible things..." E.g., 100 years ago it was indeed interesting to hear from Kuratowski, Wiener, and others, that "an ordered pair" $(a,b)$ could be modeled "in set theory" (wherein "everything's a set") as $\{\{a\}, \{a,b\}\}$. But, srsly, folks, it is the characterization of "ordered pairs", namely, that $(a,b)=(a',b')$ if an only if $a=a'$ and $b=b'$ that we care about. That is, we care about the interaction of "an ordered pair" with the individuals "in" it, and with other ordered pairs, not the construction... although it's nice to know that clever people can model ordered pairs as sets.
Second, ... and here's a not-so-elementary point... a good "external" characterization of a thing often so-completely characterizes it that, by the very characterization (if done right), any two things $X,Y$ that do the job have a (unique) isomorphism between them (that also fits perfectly with whatever job is required). At a point where one distinguishes "equality" from "isomorphism" (which is sometimes apt, sometimes not), it is noteworthy that this does not say $X=Y$, but only that they are (uniquely) isomorphic. In truth, mostly this is the most one can reasonably demand. Often, this part follows _prior_to_ any construction! By the characterization! To repeat: although this is somewhat alien to elementary mathematics (typical undergrad?), the construction is not of interest, the interaction (as external characterization) is the point.
For example, with more-familiar objects: a quotient group $Q=G/N$ is a group $Q$ and (quotient) homomorphism $q:G\to Q$ such that any group hom $F:G\to H$ with $N$ inside its kernel "factors through $Q$, in the sense that there is $f:Q\to H$ such that $F=f\circ q$. Drawing a picture is nice, here, too.
What is an "indeterminate" $x$, as in the case of a polynomial ring $k[x]$ over a field $k$? We think of "variable" things, quite reasonable, but, seriously, what can this mean? It means that, for example, given a commutative ring $R$ with $k$ inside it, and given $a\in R$, there is a unique ("$k$-algebra") homomorphism $k[x]\to R$ sending $x$ to $a$.
For tensor products: a tensor product $V\otimes W$ of two $k$-vectorspaces is a $k$-vectorspace and bilinear $b:V\times W\to V\otimes W$ such that, for every bilinear $\beta:V\times W\to X$ for $k$-vectorspace $X$, there is a unique $k$-linear map $B:V\otimes W\to X$ through which $\beta$ factors, by $\beta=B\circ b$. From this characterization (not construction) all the relevant things can be proven. Does such a thing exist? Well, yes, and there are various (not so useful, really) constructions.
Part of the issue is that we are often given "the definition" of tensor products without any persuasive argument for caring about such things... and, then, the fact that they are somewhat more sophisticated things makes it doubly confusing.
So, that was the five-minute version of something. :)