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I'm mostly self taught in undergraduate physics and maths, so I haven't had much of an education in this stuff. Basically, I know how the theory of general relativity is formulated in terms of tensors and tensor equations. You equate components of the curvature tensor with components of the energy-momentum tensor, as in $R_{\mu \nu} -\frac{1}{2}Rg_{\mu \nu} = \frac{8\pi G}{c^4}T_{\mu \nu}$. But I wasn't really satisfied with this, because of how much it seemed to depend on the components of the tensor. Hence I started to research other definitions of tensors (after all, vectors can be defined completely independently of components).

What I found was that tensors are defined as multilinear maps: \begin{align} T: V^*\times \cdots \times V^*\times V\times \cdots \times V\rightarrow \mathbb{R}.\end{align}

Now I understand all of the terms in this definition (e.g. dual spaces, direct products), but I don't know how this relates to the definition I've seen before in physics, as objects that are invariant under coordinate transformations: \begin{align}(T')^{m_1 \cdots} _{n_1 \cdots} = \frac{\partial (x')^{m_1}}{\partial x^{p_1}}\cdots \frac{\partial x^{q_1}}{\partial (x')^{n_1}}\cdots T^{p_1 \cdots}_{q_1 \cdots}.\end{align} So my question is this: how do these definitions relate to each other? They seem completely different, and I'm not even sure how to think of a "physics tensor" as a multilinear map to $\mathbb{R}$. Also, if you're knowledgeable in physics, is there a form of the Einstein Field Equations that is completely independent of coordinates?

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    $\begingroup$ Once you "unpick" a basis EFE can be rewritten as $$\overleftrightarrow{R} - \frac{1}{2}R\overleftrightarrow{g} = 8\pi \overleftrightarrow{T}$$ which is a coordinate independent equation between geometric objects. $\endgroup$ Commented May 14, 2020 at 8:52
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    $\begingroup$ I think one point of confusion is that physicists tend to say "tensor" when they really mean "tensor field". (Is that right?) I looked at Sean Carroll's book Spacetime and Geometry and it appeared to begin with an introduction to manifolds that seemed consistent with explanations I've seen in math textbooks; so perhaps that book is helpful to read. $\endgroup$
    – littleO
    Commented May 14, 2020 at 8:55

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To simplify notation, let me restrict the discussion to $(1,1)$-tensors. Unfortunately, one uses the word "tensor" in mathematics in (at least) two different contexts:

  1. In multilinear algebra, a $(1,1)$-tensor can be indeed defined as a multilinear map $T \colon V^{*} \times V^{*} \rightarrow \mathbb{R}$. This is a "coordinate-free" definition.
  2. In differential geometry (which is the relevant framework for general relativity) and physics, given a manifold $M$, a $(1,1)$-tensor $\mathcal{T}$ on $M$ is a map which gives you at each point $p \in M$ a multilinear map $\mathcal{T}|_p \colon \left( T_p M \right)^{*} \times T_pM \rightarrow \mathbb{R}$ which varies smoothly with $p$. Such a map is sometimes called a "tensor field" because it isn't really a tensor on a vector space in the sense of multilinear algebra but a whole family of tensors, one for each $p \in M$ and the tensor you get at a point $p$ is defined on a completely different vector space than the tensor you get at a point $q \neq p$. Again, this is a coordinate-free definition.

How is this related to the transformation rule you wrote?

  1. By choosing a basis $(v_1,\dots,v_n)$ for $V$ and taking the dual basis $(v^1,\dots,v^n)$ for $V^{*}$, a tensor $T \colon V^{*} \times V \rightarrow \mathbb{R}$ can be encoded by a collection of numbers $T^i_j = T(v^i,v_j)$ (when $1\leq i,j\leq n$). The collection of numbers will depend on the choice of basis (i.e, the coordinates). If you have a different basis $(w_1,\dots,w_n)$ which is related to the original basis by $w_j = S_j^k v_k$ then $w^j = \left( S^{-1} \right)^j_k v^k$ and so $$ (T')^i_j = T(w^i, w_j) = T \left( \left( S^{-1} \right)^i_k v^k, S_j^l v_l \right) = \left( S^{-1} \right)^i_k S_j^l T(v^k,v_l) = \left( S^{-1} \right)^i_k S_j^l T^k_l. $$ This is the transformation rule between the representation of a multilinear map in two different bases.
  2. By choosing a coordinate system $(x^1,\dots,x^n)$ around $p \in M$, you get bases $(\partial_{x_1}, \dots, \partial_{x_n})$ for all the tangent spaces $T_pM$ around $p$ and the corresponding dual bases $(dx^1, \dots, dx^n)$. Then you can represent your tensor field $\mathcal{T}$ at each $p$ by the collection of numbers $\mathcal{T}^i_j = \mathcal{T} \left( dx^i, \partial_{x_j} \right)$. What you get is a collection of $n^2$ functions, not numbers because you do it for each point in your neighborhood. What happens when you choose a different coordinate system $((x')^1, \dots, (x')^n)$ around $p$? You get different bases $(\partial_{x'_1}, \dots, \partial_{x'_n})$ for the tangent spaces $T_pM$, different dual bases $(d(x')^1,\dots,(dx')^n)$ and different numbers $\left( \mathcal{T}'\right)^i_j = \mathcal{T}(d(x')^i,\partial_{x'_j})$. The relation between the two bases is given by $$ \partial_{x'_j} = \frac{\partial x^k}{\partial (x')_j} \partial_{x_k}, \,\,\, d(x')^j = \frac{\partial (x')^j}{\partial x'_k} dx^k $$ and if you plug it into the formula in $(1)$, you get $$ \left( \mathcal{T}'\right)^i_j = \frac{\partial (x')^i}{\partial x'_k} \frac{\partial x^l}{\partial (x')_j} T^k_l $$ which is exactly the formula you quote.

In physics, one usually doesn't start with the mathematical description of tensors and tensor fields I just gave you and instead they usually take the transformation rule as the "definition" of a tensor. More formally, physicists usually think of tensors as a rule which assigns for each coordinate system functions $\mathcal{T}^i_j$ such that for different coordinate systems, the components of the functions are related by the transformation rule above. This avoids all sorts of mathematical discussions about multi-linearity, tensor products, tensor bundles but can obscure what a "is really" a tensor.

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    $\begingroup$ Tangential comment, but in applied math (such as in the book Golub and Van Loan) and machine learning the word "tensor" often means simply a multidimensional array of numbers. So that's a third use of the word "tensor". I mention this just in case anyone is wondering what machine learning people mean when they say "tensor". $\endgroup$
    – littleO
    Commented May 14, 2020 at 21:35

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