At least in my undergrad multivariate real analysis class, I remember the professor arranging things to strongly suggest that the Hessian should be thought of as ∇⊗∇, and that this was the second term in a higher dimensional Taylor series, so that the third derivative term would be ∇⊗∇⊗∇ etc. Things like tensor products or even quotient spaces weren't assumed knowledge, so it wasn't explicitly covered, but I remember feeling the connection was obvious enough at the time. Then an introductory differential geometry class got into (n,m) tensors. So I'm quite sure mathematicians are fine dealing with tensors. My experience was undergrad engineering math tries to avoid even covectors though, so that will stay well clear of a coherent picture of multi-variable calculus. e.g. my engineering professors would talk of dirac δ as an infinite spike/spooky doesn't-really-exist thing that makes integrals work or whatever. My analysis professor just said δ(f) = f(0) is a linear functional.

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tho2342i342342 3 days ago | parent [-]

∇⊗∇ would be more like \p_i f . \p_j f, not \p_{ij} f

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setopt 3 days ago | parent [-]

I disagree, if you apply it in the order (∇⊗∇) f then you should get \partial_i \partial_j as elements of a rank-2 operator that is then applied to a function f. That is, presumably, what you mean by \p_{ij} f.

	▲	ndriscoll 3 days ago \| parent \| next [-]
		I think it is actually bad notation. It was a long time ago but I think the suggestive nod was basically pointing out how the entries of the matrix are arranged as "copies of del" and that this thing should be thought of as eating 2 infinitesimal vectors. I don't think he actually wrote that (again, the prereqs didn't make sense to try to do so), but afaik some people do use that notation. I think if you tried to actually make sense of it, you'd expect that you plug in f twice (I think ∇ already secretly lives in R^n* ⊗ L(C^infty, C^infty), so you'd expect 2 copies of each of those). If you think of it as ∇: L(C^infty, L(R^n, C^infty)) then the composition ∇^2: L(C^infty, L(R^n, L(R^n, C^infty))) ≅ C^infty* ⊗ R^n* ⊗ R^n* ⊗ C^infty, so the types should work correctly (though the type of ∇ actually needs some generic parameters for composition to make sense), and you get that you only plug f in once. Unfortunately ∇^2 is already taken by the Laplacian, so I suppose ∇⊗∇ would be "∇^2, but the one that makes a tensor", and it'll do as long as you don't try to type check it, which physicists won't.
	▲	3 days ago \| parent \| prev [-]
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