# Definition of tensors - abstract and concrete

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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.

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".

Last edited:
atyy and WWGD
Dr. Courtney
Gold Member
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.

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