Csdrhrt

I am an undergraduate student of Physics, currently taking a course on Special Relativity, but I am getting too confused with tensors and their indices.

My question is: How to convert mixed tensors to contravariant or covariant tensors, and is it possible to interchange indices in a mixed tensor?

A $(p, q)$-tensor on a real vector space $V$ is a multilinear map $T : (V^*)^ptimes V^q to mathbb{R}$.

Let ${e_1, dots, e_n}$ be a basis for $V$ and ${e^1,dots, e^n}$ the dual basis of $V^*$, then the tensor $T$ is determined by the collection of real numbers $T^{i_1, dots, i_p}_{j_1, dots, j_q} := T(e^{i_1},dots, e^{i_p}, e_{j_1}, dots, e_{j_q})$. If ${hat{e}_1, dots, hat{e}_n}$ is another basis for $V$ and ${hat{e}^1, dots, hat{e}^n}$ is the corresponding dual basis, then we get another collection of real numbers $hat{T}^{i_1',dots, i_p'}_{j_1', dots, j_q'} := T(hat{e}^{i_1'}, dots, hat{e}^{i_p'}, hat{e}_{j_1'},dots, hat{e}_{j_q'})$.

If $A$ denotes the change of basis matrix from ${e_1, dots, e_n}$ to ${hat{e}_1, dots, hat{e}_n}$ then, using the Einstein summation convention, we have $hat{e}_i = A^k_ie_k$. The change of basis matrix from ${e^1, dots, e^n}$ to ${hat{e}^1, dots, hat{e}^n}$ is $A^{-1}$ so $hat{e}^j = (A^{-1})^j_k e^k$. It follows that

$$hat{T}^{i_1',dots,i_p'}_{j_1',dots,j_q'} = T^{i_1,dots,i_p}_{j_1,dots,j_q}(A^{-1})^{i_1'}_{i_1}dots(A^{-1})^{i_p'}_{i_p}A^{j_1}_{j_1'}dots A^{j_q}_{j_q'}.$$

In physics, a $(p, q)$-tensor is often considered as a collection of real numbers $T^{i_1,dots, i_p}_{j_1,dots, j_q}$ which transforms under change of basis in the way stated above. As the indices $j_1, dots, j_q$ change according to the change of basis matrix, we say that they are covariant, while the indices $i_1, dots, i_p$ change according to the inverse of the change of basis matrix, so we say that they are contravariant. Hence a $(p, q)$-tensor has $p$ contravariant indices and $q$ covariant indices.

Examples:

• A $(0, 1)$-tensor is nothing but a linear map $V to mathbb{R}$.


• Given a vector $v in V$, one obtains a $(1, 0)$-tensor $T_v$ defined by $T_v(alpha) = alpha(v)$.


• An inner product on $V$ is an example of a $(0, 2)$-tensor.


• A linear map $L : V to V$ can be viewed as a $(1, 1)$-tensor $T_L$ defined by $T_L(alpha, v) = alpha(L(v))$.

A (not necessarily positive-definite) inner product $g$ defines an isomorphism $Phi_g : V to V^*$ given by $Phi_g(v) = g(v, cdot)$. This isomorphism can be used to transform a $(p, q)$-tensor $T$ into a $(p - 1, q + 1)$-tensor $T'$ by defining $T'(alpha^1, dots, alpha^{p-1}, v_1, dots, v_{q+1}) := T(alpha^1, dots, alpha^{p-1}, Phi_g(v_1), v_2, dots, v_{q+1})$. Likewise, the inverse isomorphism $Phi_g^{-1}$ can be used to transform a $(p, q)$-tensor into a $(p + 1, q - 1)$-tensor. Doing this repeatedly, we can view a $(p, q)$-tensor as an $(r, s)$-tensor for any $r$ and $s$ with $r, s geq 0$ and $r + s = p + q$. Note however that the $(r, s)$-tensor we produce depends on the inner product $g$; for a different inner product, the corresponding $(r, s)$-tensor will not be the same.

A $(p, q)$-tensor field on a smooth manifold $M$ is $C^{infty}(M)$ multilinear map $T : Gamma(T^*M)^ptimesGamma(TM)^q to C^{infty}(M)$. That is, a $(p, q)$-tensor on $T_xM$ for every $x in M$ which varies smoothly as $x$ varies.

Given local coordinates $(x^1, dots, x^n)$ on $U subseteq M$, there is a basis of sections for $TM|_U$ given by ${partial_1, dots, partial_n}$ where $partial_i = frac{partial}{partial x^i}$, and a dual basis of sections for $T^*M|_U$ given by ${dx^1, dots, dx^n}$. We then obtain a collection of smooth functions $T^{i_1,dots,i_p}_{j_1,dots,j_q} := T(dx^{i_1},dots, dx^{i_p}, partial_{j_1}, dots, partial_{j_q})$ on $U$. If ${hat{x}^1, dots, hat{x}^n}$ is another set of local coordinates on $U$, then ${hat{partial}_1, dots, hat{partial}_n}$ is a basis of sections for $TM|_U$ where $hat{partial}_i = frac{partial}{partialhat{x}^i}$, and ${dhat{x}^1,dots, dhat{x}^n}$ is the dual basis of sections for $T^*M|_U$, so we get another collection of smooth functions $hat{T}^{i_1',dots,i_p'}_{j_1',dots,j_q'} := T(dhat{x}^{i_1'},dots, dhat{x}^{i_p'}, hat{partial}_{j_1'},dots, hat{partial}_{j_q'})$ on $U$.

Note that $hat{partial}_i = dfrac{partial x^k}{partial hat{x}^i}partial_k$ and $dhat{x}^j = dfrac{partial hat{x}^j}{partial x^k}dx^k$ so

$$hat{T}^{i_1',dots,i_p'}_{j_1',dots,j_q'} = T^{i_1,dots,i_p}_{j_1,dots,j_q}dfrac{partial hat{x}^{i_1'}}{partial x^{i_1}}dots dfrac{partial hat{x}^{i_p'}}{partial x^{i_p}}dfrac{partial x^{j_1}}{partial hat{x}^{j_1'}}dots dfrac{partial x^{j_q}}{partial hat{x}^{j_q'}}$$

Recall that $left(dfrac{partialhat{x}}{partial x}right)^{-1} = dfrac{partial x}{partialhat{x}}$, so the above is completely analogous to the previous formula for tensors.

Examples:

• A $(0, 1)$-tensor field is nothing but a one-form.


• Given a vector field $V in Gamma(TM)$, one obtains a $(1, 0)$-tensor field $T_V$ defined by $T_V(alpha) = alpha(V)$.


• A Riemannian or Lorentzian metric on $M$ is an example of a $(0, 2)$-tensor field.


• A bundle map $L : TM to TM$ can be viewed as a $(1, 1)$-tensor $T_L$ defined by $T_L(alpha, V) = alpha(L(V))$.

As in the tensor case, given a Riemannian or Lorentzian metric (or a non-degenerate metric of any signature), one can transform a $(p, q)$-tensor field into a $(r, s)$-tensor field for any $r, s geq 0$ with $r + s = p + q$.