Tensor product of vector bundles

Question

The Whitney sum (where fibre dimensions add) of two real, or two complex, vector bundles $\pi : E \to X$ and $\pi' : E' \to X$ over a topological space $X$ is not hard to get an intuitive grasp of. Especially if you consider the tangent bundle of a smooth manifold embedded in Euclidean space and its normal bundle. (I'm mainly interested in the cases where $X$ is a smooth manifold.)

But I have never felt I had much intuition for what is going on geometrically when one takes the tensor product (where fibre dimensions multiply) of the bundles $\pi$ and $\pi'$, even though the definition is entirely familiar.

Perhaps the most interesting case is the Picard group of equivalence classes of complex line bundles over a complex manifold, under the operation of tensor product.

Can someone offer some geometrical intuition about what is going on when we take the tensor product of real or complex vector bundles?

Answer 1

Perhaps the fundamental issue here is an unintended consequence of the orthodox set-theoretic foundations of mathematics, which can give the mistaken impression that mathematical objects need be encoded as sets in order to do math "properly", even if this comes at the cost of being able to intuitively understand these objects.

Let me explain (and be patient, I will get to tensor products of vector bundles eventually). We are all familiar for instance with the concept of a function $f: X \to Y$, but strictly speaking a function is not a set and so does not directly exist inside a standard set theory such as ZFC. But we have a standard solution to this: we encode a function $f$ within set theory by identifying a function with its graph $\Gamma_f := \{ (x,f(x)): x \in X \}$ (or, if one wants to be more pedantic, with the ordered triple $(X, Y, \Gamma_f)$ in order to record the domain $X$ and codomain $Y$, with the ordered triple itself being encoded as a set in some standard fashion). Every function uniquely determines a graph and every graph uniquely determines a function (if we also specify the domain and codomain), so at a foundational level this is a satisfactory resolution to the problem of introducing functions into set theory. But it comes at the cost of intuition. Consider for instance the problem of trying to define the pointwise product $fg: X \to {\bf R}$ of two real-valued functions $f,g: X \to {\bf R}$. If we view a real-valued function $f: X \to {\bf R}$ as a real number $f(x)$ that is parameterized by some parameter $x \in X$, then the pointwise product is simply the ordinary multiplication operation on the real numbers, with the only difference being that the real numbers involved are not constant, but instead depend on a parameter $x$: $$ fg(x) := f(x) g(x).$$ This is extremely intuitive (especially if $x$ is interpreted as a time variable, a position variable, or a state in a state space or probability space). For instance, it is immediately obvious that all the usual commutative ring axioms for real numbers extend immediately to real-valued functions. However, if one insists on interpreting functions $f$ as graphs $\Gamma_f$, the definition suddenly becomes horrible: $$ \Gamma_{fg} := \{ (x,c): \exists a,b \in {\bf R}: (x,a) \in \Gamma_f, (x,b) \in \Gamma_g, ab=c \}.$$ If one insists on the graph-based encoding as the "geometric" way to think about functions, then the operation of multiplying two functions together is now much more opaque than it needs to be. For example, it is not obvious at all that this operation is associative.

The situation with tensor products (or many other standard operations) of vector bundles is completely analogous. In our orthodox set-theoretic foundations, we define a vector bundle over a base $X$ as a set $E$ equipped with various structures (a projection map $\pi: E \to X$, a smooth structure on $E$, etc.) obeying various axioms. But one can instead think of a vector bundle as a vector space $E_x$ which is not constant, but instead depends (continuously, smoothly, or holomorphically) on a parameter $x \in X$. This is completely akin to viewing a function not as a graph $\Gamma_f$, but as a quantity $f(x)$ depending on a parameter $x$. In many ways, the latter interpretation is more intuitive; but it is more difficult to shoehorn into the orthodox set-theoretic foundations of mathematics, so we use the former definition instead. (See also this previous MO answer of mine for a related discussion.) So one should often think of a vector bundle as a vector space that is varying in time, or in space, or is dependent on the state of a system, or whatever other interpretation is germane for the application at hand. (Perhaps the fancy way of saying this is that vector bundles can be thought of as vector spaces in a topos over the base space.)

Anyway, once one adopts the parameterized point of view, the tensor product operation on bundles effortlessly generalizes the tensor product operation on vector spaces: $$ (E \otimes F)_x := E_x \otimes F_x.$$ So whatever intuition you have for tensor products of vector spaces, now transfers instantly to vector bundles.

Answer 2

I guess we should start by speaking somewhat philosophically about what it means to think geometrically about bundles. For me, this usually means looking at their sections: either constructing interesting sections (e.g. the unit normal field to an oriented embedded hypersurface) or proving that sections with certain properties don't exist (e.g. nonvanishing vector fields on the 2-sphere). Note that this perspective doesn't actually lose anything: the functor which sends a vector bundle to its space of sections is an equivalence between the category of vector bundles over a compact Hausdorff space X and the category of finitely generated projective modules over $C(X)$.

So if we're willing to shift our focus from the bundle to the sections, it becomes clear that we think geometrically about tensor products all the time: for instance a Riemannian metric over a manifold $M$ is just a section of $(TM \otimes TM)^*$ with certain properties. Likewise the Riemannian curvature tensor is a section of

$$T^*M \otimes T^*M \otimes Hom(TM, TM)$$

It's admittedly a little hard to think straight about these bundles as bundles, but their spaces of sections contain subspaces which are easier to digest, like the space of Riemannian metrics or the space of curvature tensors. Basically any time you want to do linear algebra - bilinear forms, linear maps, eigenspace decompositions, etc. - parametrized by a base space, you have to work in a tensor bundle over that space.

Answer 3

A rank-$n$ vector bundle over $X$ consists of a bunch of trivial vector bundles over open subsets $U_i$ covering $X$, patched together on the overlaps. On each overlap $U_{ij}=U_i\cap U_j$, there is a map $U_{ij}\rightarrow GL_n(k)$ that describes the patching. (This collection of maps is not, in general, unique.)

Given two rank-$n$ vector bundles $E\rightarrow X$ and $F\rightarrow X$, you can choose a single collection $\{U_i\}$ of open sets on which both bundles are trivialized. The tensor product $E\otimes F$ is described by the tensor product of the patching maps.

In particular, a line bundle is a collection of trivial line bundles that get twisted and then patched together. The tensor product of line bundles $E$ and $F$ is constructed by twisting once (as you would twist to get $E$), then twisting again (as you would twist to get $F$) and then patching.

More succinctly, a line bundle is an element of $H^1(X,GL_1(k))$ (where $H$ is Cech cohomology and $k$ is ${\mathbb R}$ for real vector bundles, ${\mathbb C}$ for complex vector bundles. Tensor product of vector bundles corresponds to multiplication in Cech cohomology.