# Can contraction of a tensor be defined without using coordinates?

1. Aug 31, 2013

### andrewkirk

All but one of the tensor operations can be defined without reference to either coordinates or a basis. This can be done for instance by defining a $(^m_n)$ tensor over vector space $V$ as a multi-linear function from $V^m(V^*)^n$ to the background field $F$.

This allows us to define tensor addition and multiplication, and the application of a $(_n^m)$ tensor to a sequence of $r$ vectors and $s$ one-forms, where $0\leq r\leq m$ and $0\leq s\leq n$, without referring to coordinates or using any basis of $V$. For instance, for $(_n^m)$ tensors $T1$ and $T2$, the $(_n^m)$ tensor $(T1+T2):V^m(V^*)^n\rightarrow F$ is defined by

$(T1+T2)(\vec{v_1},\vec{v_2}, ....,\vec{v_m}, \tilde{p_1},\tilde{p_2},...,\tilde{p_n}) = T1(\vec{v_1},\vec{v_2}, ....,\vec{v_m}, \tilde{p_1},\tilde{p_2},...,\tilde{p_n}) + T2(\vec{v_1},\vec{v_2}, ....,\vec{v_m}, \tilde{p_1},\tilde{p_2},...,\tilde{p_n})$

and, for $(_{n1}^{m1})$ and $(_{n2}^{m2})$ tensors $T1$ and $T2$, the tensor $(T1\otimes T2):V^{m1+m2} (V^*)^{n1+n2}\rightarrow F$ is defined by

$(T1\otimes T2)(\vec{v_1},\vec{v_2}, ....,\vec{v_{m1+m2}}, \tilde{p_1},\tilde{p_2},...,\tilde{p_{n1+n2}}) = T1(\vec{v_1},\vec{v_2}, ....,\vec{v_{m1}}, \tilde{p_1},\tilde{p_2},...,\tilde{p_{n1}}) \times T2(\vec{v_1},\vec{v_2}, ....,\vec{v_{m2}}, \tilde{p_1},\tilde{p_2},...,\tilde{p_{n2}})$

The other important tensor operation is contraction. But it seems to always be defined using coordinates. It is possible to define it not explicitly referring to coordinates but instead referring to basis vectors, but that is still basis-dependent. The proof that the multilinear operator thus defined is independent of the coordinate system or basis used to define it is short and simple, but I still find it unsatisfying that an operation that does not depend on coordinates or bases should need to be defined in terms of them.

I tried a few strategies to define contraction in a coordinate/basis-independent way, but got nowhere.

Can anybody think of a way to define contraction that does not refer to either a coordinate system or a basis?

2. Jul 9, 2015

### andrewkirk

I was looking back over this old unanswered question of mine and it occurred to me that the trace of a linear operator is the sum of its eigenvalues, and that is a coordinate-independent definition.

A trace is a contraction of a $(^2_0)$ tensor. This leads me to wonder whether one might somehow be able to generalise the coordinate-independent definition of trace to a coordinate-independent definition of a contraction, by using eigenvalues. The trick is: eigenvalues of what?

3. Jul 10, 2015

### andrewkirk

Success! Here's the definition.

Let $T:{V^*}^m\otimes V^n\to\mathbb{R}$ be a $(_n^m)$ tensor.

We define the contraction of the tensor across its $m$th contravariant and $n$th covariant arguments to be:

$\textrm{Contract}(T,m,n):{V^*}^{m-1}\otimes V^{n-1}\to\mathbb{R}$ such that

$\textrm{Contract}(T,m,n)(\vec{v}_1, ...,\vec{v}_{m-1}, \tilde{p}_1,..,\tilde{p}_{n-1}) \equiv \textrm{Trace}\big(M_T(\vec{v}_1, ...,\vec{v}_{m-1}, \tilde{p}_1,..,\tilde{p}_{n-1})\big)$

where $M_T(\vec{v}_1, ...,\vec{v}_{m-1}, \tilde{p}_1,..,\tilde{p}_{n-1}):V\to V$ such that $M_T(\vec{v}_1, ...,\vec{v}_{m-1}, \tilde{p}_1,..,\tilde{p}_{n-1})(\vec{u})$ is the map $\tilde{q}\mapsto T(\vec{v}_1, ...,\vec{v}_{m-1},\vec{u}, \tilde{p}_1,..,\tilde{p}_{n-1},\tilde{q})$

Note that the range of this map is $\mathbb{R}$, so the map
$M_T(\vec{v}_1, ...,\vec{v}_{m-1}, \tilde{p}_1,..,\tilde{p}_{n-1})(\vec{u}):V^*\to\mathbb{R}$
must itself be an element of $V$.

All we have to do is show that (1) the Trace is well-defined, and (2) $\textrm{Contract}(T,m,n)$ is linear in all its $m+n-2$ arguments.

Firstly, $M_T(\vec{v}_1, ...,\vec{v}_{m-1}, \tilde{p}_1,..,\tilde{p}_{n-1})$ (let's call it $M_T(\textrm{etc.})$ for short) is linear because $T$ is linear in all its arguments, and $M_T(\textrm{etc.})$ is $T$ with all but two arguments supplied and fixed. Hence $M_T(\textrm{etc.})$ is linear and so its trace is well-defined.

Secondly, the map $(\vec{v}_1, ...,\vec{v}_{m-1}, \tilde{p}_1,..,\tilde{p}_{n-1})\mapsto M_T(\vec{v}_1, ...,\vec{v}_{m-1}, \tilde{p}_1,..,\tilde{p}_{n-1})$ similarly inherits linearity in all its arguments from $T$.

Thirdly the `Trace' map is known to be linear. Note also that the Trace map is coordinate-independent because the trace of a linear operator is the sum of its eigenvalues.

$\textrm{Contract}(T,m,n)$ is just the composition of the second and third of these, and hence is linear as required.

We can then define the contraction of the tensor across its $i$th contravariant and $j$th covariant arguments, where $1\leq i\leq m$ and $1\leq j\leq n$, to be:

$\textrm{Contract}(T,i,j)\equiv \textrm{Contract}\bigg(T_{(j\leftrightarrow n)}^{(i\leftrightarrow m)},m,n\bigg)$

where $T_{(j\leftrightarrow n)}^{(i\leftrightarrow m)}$ is $T$ with its $i$th and $m$th contravariant arguments swapped and its $j$th and $n$th contravariant arguments swapped. That is:

$T_{(j\leftrightarrow n)}^{(i\leftrightarrow m)}(\vec{v}_1, ..,\vec{v}_{m}, \tilde{p}_1,..,\tilde{p}_{n}) \equiv T(\vec{v}_1,..,\vec{v}_{i-1},\vec{v}_m,\vec{v}_{i+1}, ..,\vec{v}_{m-1},\vec{v}_i, \tilde{p}_1,..,\tilde{p}_{j-1},\tilde{p}_{n},\tilde{p}_{j+1}..,\tilde{p}_{n-1},\tilde{p}_{j})$

It is straightforward, but somewhat long-winded, to show that the contraction defined in this way is the same as the contraction $\textrm{Contract}^\textrm{Co-ord}(T,m,n)$ defined using coordinates. It proceeds by showing that, given a basis for $V$ and its dual basis for $V^*$, Contract$(T,m,n)$ applied to any combination of $(m-1)+(n-1)$ basis elements gives the same number in $\mathbb{R}$ as is given by applying $\textrm{Contract}^\textrm{Co-ord}(T,m,n)$ to those arguments.