# Geometric intuition for the tensor product of vector spaces

First of all, I am very comfortable with the tensor product of vector spaces. I am also very familiar with the well-known generalizations, in particular the theory of monoidal categories. I have gained quite some intuition for tensor products and can work with them. Therefore, my question is not about the definition of tensor products, nor is it about its properties. It is rather about the mental images. My intuition for tensor products was never really geometric. Well, except for the tensor product of commutative algebras, which corresponds to the fiber product of the corresponding affine schemes. But let's just stick to real vector spaces here, for which I have some geometric intuition, for example from classical analytic geometry.

The direct product of two (or more) vector spaces is quite easy to imagine: There are two (or more) "directions" or "dimensions" in which we "insert" the vectors of the individual vector spaces. For example, the direct product of a line with a plane is a three-dimensional space.

The exterior algebra of a vector space consists of "blades", as is nicely explained in the Wikipedia article.

Now what about the tensor product of two finite-dimensional real vector spaces $V,W$? Of course $V \otimes W$ is a direct product of $\dim(V)$ copies of $W$, but this description is not intrinsic, and also it doesn't really incorporate the symmetry $V \otimes W \cong W \otimes V$. How can we describe $V \otimes W$ geometrically in terms of $V$ and $W$? This description should be intrinsic and symmetric.

Note that SE/115630 basically asked the same, but received no actual answer. The answer given at SE/309838 discusses where tensor products are used in differential geometry for more abstract notions such as tensor fields and tensor bundles, but this doesn't answer the question either. (Even if my question gets closed as a duplicate, then I hope that the other questions receive more attention and answers.)

More generally, I would like to ask for a geometric picture of the tensor product of two vector bundles on nice topological spaces. For example, tensoring with a line bundle is some kind of twisting. But this is still some kind of vague. For example, consider the Möbius strip on the circle $S^1$, and pull it back to the torus $S^1 \times S^1$ along the first projection. Do the same with the second projection, and then tensor both. We get a line bundle on the torus, okay, but how does it look like geometrically?

Perhaps the following related question is easier to answer: Assume we have a geometric understanding of two linear maps $f : \mathbb{R}^n \to \mathbb{R}^m$, $g : \mathbb{R}^{n'} \to \mathbb{R}^{m'}$. Then, how can we imagine their tensor product $f \otimes g : \mathbb{R}^n \otimes \mathbb{R}^{n'} \to \mathbb{R}^m \otimes \mathbb{R}^{m'}$ or the corresponding linear map $\mathbb{R}^{n n'} \to \mathbb{R}^{m m'}$ geometrically? This is connected to the question about vector bundles via their cocycle description.

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not really geometrical (and certainly not an answer), but since in the finite-dim. case $V^*\otimes W^*$ is the same as bilinear functions on $V\times W$, and since $V=(V^*)^*$, bi-(or multi-)linear forms give rather simple realization of tensor products. And at least sometimes they are what you really want from geometrical point of view. – user8268 Sep 6 '13 at 20:31
Good question, I look forward to the answers. If $W^*$ is the dual of $W$, I always feel more comfortable seeing $V\otimes W^*$ as the vector space of linear maps $L(W,V)$, and in particular the tensors $v\otimes w^*$ as the rank $1$ maps $w^*(\cdot)v$. Unfortunately this requires an identification of $W$ with $W^*$ to describe $V\otimes W$. But at least the identification $V\otimes W^*\simeq L(W,V)\simeq L(V^*,W^*)\simeq W^*\otimes V^{**}\simeq W^*\otimes V$ is intrinsic via the transpose and the canonical identification $V\simeq V^{**}$. – user85486 Sep 6 '13 at 20:32
@user8268,Humanity: These are just the usual familiar algebraic incarnations of the tensor product - nothing geometric. Of course one may also ask for a geometric interpretation of the vector space of linear maps (it is clear what a linear map is geometrically, but this doesn't describe the "shape" of the whole space of them). – Martin Brandenburg Sep 6 '13 at 21:46
I think you should have a look at thí question : math.stackexchange.com/questions/115630/… – Knumber10 Sep 16 '13 at 8:33
I don't quite understand how you accept the geometric picture of the exterior algebra, or rather in what sense it is better. Clearly, in $\Lambda^k$ we understand the decomposables as $k$-subspaces, but for me it's far from clear how to interpret geometrically all the ways to write an arbitrary $k$-vector as a "mixture" of decomposables. The same goes on here: understanding the decomposables is easy - they're just pairs of vectors - but again, arbitrary tensors are "mixtures" of these pairs... On the other hand, the 2-nd symmetric power consists of ellipsoids. :) – Alexander Shamov Oct 15 '13 at 19:50

Well, this may not qualify as "geometric intuition for the tensor product", but I can offer some insight into the tensor product of line bundles.

A line bundle is a very simple thing -- all that you can "do" with a line is flip it over, which means that in some basic sense, the Möbius strip is the only really nontrivial line bundle. If you want to understand a line bundle, all you need to understand is where the Möbius strips are.

More precisely, if $X$ is a line bundle over a base space $B$, and $C$ is a closed curve in $B$, then the preimage of $C$ in $X$ is a line bundle over a circle, and is therefore either a cylinder or a Möbius strip. Thus, a line bundle defines a function $$\varphi\colon \;\pi_1(B)\; \to \;\{-1,+1\}$$ where $\varphi$ maps a loop to $-1$ if its preimage is a Möbius strip, and maps a loop to $+1$ if its preimage is a cylinder.

It's not too hard to see that $\varphi$ is actually a homomorphism, where $\{-1,+1\}$ forms a group under multiplication. This homomorphism completely determines the line bundle, and there are no restrictions on the function $\varphi$ beyond the fact that it must be a homomorphism. This makes it easy to classify line bundles on a given space.

Now, if $\varphi$ and $\psi$ are the homomorphisms corresponding to two line bundles, then the tensor product of the bundles corresponds to the algebraic product of $\varphi$ and $\psi$, i.e. the homomorphism $\varphi\psi$ defined by $$(\varphi\psi)(\alpha) \;=\; \varphi(\alpha)\,\psi(\alpha).$$ Thus, the tensor product of two bundles only "flips" the line along the curve $C$ if exactly one of $\varphi$ and $\psi$ flip the line (since $-1\times+1 = -1$).

In the example you give involving the torus, one of the pullbacks flips the line as you go around in the longitudinal direction, and the other flips the line as you around in the meridional direction:

Therefore, the tensor product will flip the line when you go around in either direction:

So this gives a geometric picture of the tensor product in this case.

Incidentally, it turns out that the following things are all really the same:

1. Line bundles over a space $B$

2. Homomorphisms from $\pi_1(X)$ to $\mathbb{Z}/2$.

3. Elements of $H^1(B,\mathbb{Z}/2)$.

In particular, every line bundle corresponds to an element of $H^1(B,\mathbb{Z}/2)$. This is called the Stiefel-Whitney class for the line bundle, and is a simple example of a characteristic class.

Edit: As Martin Brandenburg points out, the above classification of line bundles does not work for arbitrary spaces $B$, but does work in the case where $B$ is a CW complex.

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Thank you. Finally some pictures! Your descriptions of line bundles don't hold in general, you have to assume that $B$ is a CW-complex for example. – Martin Brandenburg Nov 11 '13 at 9:28
More generally, $n$-dimensional vector bundles on an arbitrary ringed space are classified by Cech coycles in $\mathrm{GL}_n(\mathcal{O}_X)$, and the tensor product corresponds to the usual multiplication on $\mathrm{GL}_n(\mathcal{O}_X)$. You have described the case $n=1$ and $X$ paracompact with $\mathcal{O}_X =$ continuous functions. The cocycle description is useful for calculations, but not really instrinsic. – Martin Brandenburg Nov 11 '13 at 9:45
@MartinBrandenburg That's a good point about the requirements on $B$ -- I will add this to my post. Also, while it's true that $n$-dimensional vector bundles correspond to Cech cocycles, the tensor product doesn't always correspond to multiplication: in dimensions greater than one, you have to take the matrix tensor product of a $\mathrm{GL}_m(\mathcal{O}_X)$ cocycle and a $\mathrm{GL}_n(\mathcal{O}_X)$ cocycle to get a $\mathrm{GL}_{mn}(\mathcal{O}_X)$ cocycle. – Jim Belk Nov 11 '13 at 16:08
Ah, yes. Also known as Kronecker product. – Martin Brandenburg Nov 11 '13 at 19:49

Good question. My personal feeling is that we gain true geometric intuition of vector spaces only once norms/inner products/metrics are introduced. Thus, it probably makes sense to consider tensor products in the category of, say, Hilbert spaces (maybe finite-dimensional ones at first). My geometric intuition is still mute at this point, but I know that (for completed tensor products) we have an isometric isomorphism $$L^2(Z_1) \otimes L^2(Z_2) \cong L^2(Z_1 \times Z_2)$$
where $Z_i$'s are measure spaces. In the finite-dimensional setting one, of course, just uses counting measures on finite sets. From this point, one can at least rely upon analytic intuition for the tensor product (Fubini theorem and computation of double integrals as iterated integrals, etc.).

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This is quite similar to the natural isomorphism $\Gamma(X) \otimes_{\Gamma(S)} \Gamma(Y) \cong \Gamma(X \times_S Y)$ for affine schemes $X,Y,S$, and also $C(X) \hat{\otimes} C(Y) \cong C(X \times Y)$ for compact Hausdorff spaces $X,Y$. But your isomorphism just says that if $B$ (resp. $C$) is a (orthonormal) basis of $V$ (resp. $W$), then $B \times C$ is (or can be regarded as) a (orthonormal) basis of $V \otimes W$. I want to have a description which doesn't choose bases. – Martin Brandenburg Sep 7 '13 at 8:22
@MartinBrandenburg: I agree on similarity and disagree about bases. I do not need a basis for this isomorphism, but I do need an inner product (as I said, to me, "geometry" does not make much sense without a metric). – studiosus Sep 7 '13 at 13:08
I wanted to say that an arbitrary Hilbert space is not given as $L^2$, this is exactly the choice of an orthonormal basis. – Martin Brandenburg Sep 8 '13 at 14:08