# How are the different definitions of a tensor the same definition?

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?

• 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. Commented May 14, 2020 at 8:52
• 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. Commented May 14, 2020 at 8:55

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.

• 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". Commented May 14, 2020 at 21:35