# Definition of tensors - abstract and concrete

#### hunt_mat

Homework Helper
I am well aware of an abstract definition of a general tensor as a map:
$$\mathbf{T}:\overbrace{V\times\cdots\times V}^{n}\times\underbrace{V^{\star}\times \cdots\times V^{\star}}_{m}\longrightarrow\mathbb{R}$$
I am happy with this definition, it makes a lot of sense to me. However, the physics definition is that of transformations between co-ordinates of the coefficients of the tensor. I can't quite figure out how these two definitions match up.

Any suggestions?

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#### ShayanJ

Gold Member
Consider two bases $e_a$ and $e'_a$ related by the transformation $e'_a=P^b_{\ a} e_b$ and $e'^a=(P^{-1})^a_{\ b}e^b$.
The components of a type (k,l) tensor in the primed coordinates are given by $T'^{c_1...c_k}_{d_1..d_l}=T(e'^{c_1},...,e'^{c_k},e'_{d_1},...,e'_{d_l})$. Now if you substitute the transformations in the arguments and use the multi-linearity of T to take $P$ and $P^{-1}$ outside, you get the transformation of T.

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#### Fredrik

Staff Emeritus
Gold Member
In differential geometry, that vector space V is the tangent space at some point p in a manifold M. It's usually denoted by $T_pM$. If $x:U\to\mathbb R^n$ is a coordinate system such that $p\in U$, then the n-tuple $\big(\frac{\partial}{\partial x^1}\big|_p,\dots,\frac{\partial}{\partial x^n}\big|_p\big)$ is an ordered basis for $T_pM$. The components of this n-tuple are defined by
$$\frac{\partial}{\partial x^i}\bigg|_p f=(f\circ x^{-1})_{,i}(x(p))$$ for all $i\in\{1,\dots,n\}$ and all smooth $f:M\to\mathbb R$. (I'm using the notation $g_{,i}$ for the $i$th partial derivative of a function $g$).

The dual space of $T_pM$ is called the cotangent space of M at p. I'll denote it by $T_pM^*$. The ordered basis for $T_pM^*$ that's dual to $\big(\frac{\partial}{\partial x^1}\big|_p,\dots,\frac{\partial}{\partial x^n}\big|_p\big)$ is $\big(\mathrm dx^1|_p,\dots,\mathrm dx^n|_p\big)$. The $x^i$ are the component functions of $x$, i.e. for each $i\in\{1,\dots,n\}$, $x^i$ is the map that takes a point $q$ in $U$ to the $i$th component of the n-tuple $x(q)$. The $\mathrm d$ notation is defined by
$$\mathrm df(v)=v(f)$$ for all smooth $f:M\to\mathbb R$ and all $v\in T_pM$.

A change of coordinate system $x\to y$ induces a change of the ordered basis and its dual:
\begin{align*}
\left(\frac{\partial}{\partial x^1}\bigg|_p,\dots,\frac{\partial}{\partial x^n}\bigg|_p\right) &\to \left(\frac{\partial}{\partial y^1}\bigg|_p,\dots,\frac{\partial}{\partial y^n}\bigg|_p\right)\\
\left(\mathrm dx^1|_p,\dots,\mathrm dx^n|_p\right) &\to \left(\mathrm dy^1|_p,\dots,\mathrm dy^n|_p\right).
\end{align*} Because of this, the coordinate change also induces a change of the components of a tensor. For example, the induced change of the ${}^i{}_{jk}$ component of a tensor $T:T_pM^*\times T_pM\times T_pM\to\mathbb R$ is
$$T\left(\mathrm dx^i|_p,\frac{\partial}{\partial x^j}\bigg|_p, \frac{\partial}{\partial x^k}\bigg|_p\right)\to T\left(\mathrm dy^i|_p,\frac{\partial}{\partial y^j}\bigg|_p, \frac{\partial}{\partial y^k}\bigg|_p\right).$$ It's not too hard to show that the relationship between the right-hand side and the left-hand side is given by the "tensor transformation law".

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#### Dr. Courtney

Gold Member
2018 Award
A Tensor is a mathematical object that obeys the tensor transformation laws.

Definitions that use formulas are expressing the tensor transformation laws in a given notation.

#### FactChecker

Gold Member
2018 Award
The physics definition of a tensor emphasizes that the tensor has the same meaning, regardless of what coordinate system and units are used. We know that forces, motion, etc exist even if no one has specified a coordinate system. In the physics definition, if two different coordinate systems are used to measure/define the same tensor, the measurements must be related in the correct way. So the tensor is really an equivalence class of all the possible ways of measuring the same physical entity in different coordinate systems and units. In that sense, the physics definition is a coordinate system agnostic definition.

The abstract definition of tensor is also a coordinate system agnostic definition. It only talks about mappings from a vector space, without mentioning a coordinate system or units in the vector spaces at all. So the abstract definition can be applied to physical entities like forces, motion, density, etc. that exist and are the same regardless of the coordinate system used to measure/define them. In that way, the abstract definition gives a simple coordinate system agnostic definition by avoiding any reference to coordinate systems, but it is also more difficult to appreciate what it is about.

In physics, things get measured in coordinate systems. In abstract math, they can just exist.

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