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> For mathematicians, a tensor is a function F. We pass the function an ordered tuple of N vectors and M covectors (or, to keep things simple, N column vectors

by JadeNB 5y ago

> For mathematicians, a tensor is a function F. We pass the function an ordered tuple of N vectors and M covectors (or, to keep things simple, N column vectors and M row vectors), and the function returns a scalar. The function F needs to be linear in each of the N vectors and M covectors. In this view, a matrix is a tensor with N=1, M=1. The operations used by machine learning types arise naturally once you crank the mathematical handle a little bit. As with almost all non-technical statements one can make about math (including my amendment—recursive nerd-sniping away), this is neither quite true nor quite false. A tensor is an element of a tensor product—that's it. (But what tensor product? There are a lot of notions. I'll use the bare tensor product of non-topologised vector spaces.) One kind of thing you can tensor is copies of a single vector space, and/or its dual. This is the sort of tensor product you likely have in mind. Tensoring copies of the dual allows you to feed in elements of the original vector space. Tensoring copies of the vector space allows you to feed in elements of the dual vector space (there is mathematically no intrinsic difference between a vector and a co-vector; having a favourite vector space in mind, you can talk about the elements of that vector space or of its dual, and—which is where the vector vs. co-vector terminology in physics comes from—about how coefficient vectors change when you change the basis of V), although, importantly, if your original vector space is infinite dimensional then the natural embedding into its double dual is not an isomorphism. Which is to say, yes, the objects you describe all arise as mathematical tensors; but mathematical tensors can also describe much more.

soVeryTired 5y ago
> if your original vector space is infinite dimensional then the natural embedding into its double dual is not an isomorphism Ok, so this feels like the crux of the matter. So how does the fact that V** is not isomorphic to V make the tensor product construction a more general concept than the linear function construction?
- JadeNB 5y ago
  > Ok, so this feels like the crux of the matter. So how does the fact that V* is not isomorphic to V make the tensor product construction a more general concept than the linear function construction? This doesn't really have anything to do with tensor products per se; it can already be seen with tensor products involving only a single factor, which are just vector spaces. Thinking of tensors only as functions means that, for example, one can never think of the original vector space V itself, only of its image in the double dual V^{**} (a fancy way of saying that you can evaluate a vector v \in V on an element v^* of the dual vector space V^* by evaluating v^* at v: in confusing but suggestive notation, v(v^*) = v^*(v)). It's certainly true that you can do this, and, given the axiom of choice, you don't lose any information; you know everything about a vector v \in V by knowing its value on elements of V^* (which is to say, by knowing the values of elements of V^* on V). However, if V is infinite dimensional, then you are forcing yourself to carry around extra, possibly unwanted information: if you are taking bare algebraic duals, not topological duals, then V^{**} is inconceivably larger than V, which is to say that there are way more linear functionals on V^* than just those coming from evaluation at a fixed element of V. You can fix some of this inconceivable largeness by knowing a little more structure carried by V, and by forcing your dual to reflect that structure—usually you know the topology, and ask that the dual consist of continuous functionals; and once there's topology, you start asking things of the tensor product, too. (For example, you probably don't want to take the vector-space tensor product of Hilbert spaces, but rather its completion in some suitable sense.) But, even with a more refined notion of duality, it's still only the nice spaces V that are identified with their double duals via the canonical map V \to V^{**}; the terminology is 'reflexive'. (I think I caught all the asterisks that the Markdown parser ate the first time through.)

by JadeNB 5y ago

soVeryTired 5y ago

> if your original vector space is infinite dimensional then the natural embedding into its double dual is not an isomorphism Ok, so this feels like the crux of the matter. So how does the fact that V** is not isomorphic to V make the tensor product construction a more general concept than the linear function construction?

JadeNB 5y ago
> Ok, so this feels like the crux of the matter. So how does the fact that V* is not isomorphic to V make the tensor product construction a more general concept than the linear function construction? This doesn't really have anything to do with tensor products per se; it can already be seen with tensor products involving only a single factor, which are just vector spaces. Thinking of tensors only as functions means that, for example, one can never think of the original vector space V itself, only of its image in the double dual V^{**} (a fancy way of saying that you can evaluate a vector v \in V on an element v^* of the dual vector space V^* by evaluating v^* at v: in confusing but suggestive notation, v(v^*) = v^*(v)). It's certainly true that you can do this, and, given the axiom of choice, you don't lose any information; you know everything about a vector v \in V by knowing its value on elements of V^* (which is to say, by knowing the values of elements of V^* on V). However, if V is infinite dimensional, then you are forcing yourself to carry around extra, possibly unwanted information: if you are taking bare algebraic duals, not topological duals, then V^{**} is inconceivably larger than V, which is to say that there are way more linear functionals on V^* than just those coming from evaluation at a fixed element of V. You can fix some of this inconceivable largeness by knowing a little more structure carried by V, and by forcing your dual to reflect that structure—usually you know the topology, and ask that the dual consist of continuous functionals; and once there's topology, you start asking things of the tensor product, too. (For example, you probably don't want to take the vector-space tensor product of Hilbert spaces, but rather its completion in some suitable sense.) But, even with a more refined notion of duality, it's still only the nice spaces V that are identified with their double duals via the canonical map V \to V^{**}; the terminology is 'reflexive'. (I think I caught all the asterisks that the Markdown parser ate the first time through.)