# Conversion of mixed tensors into mixed tensors and into covariant (or contravariant) ones

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?

• Could you provide an example? Also could you explain what you mean by a mixed tensor? – Michael Albanese Dec 17 '18 at 20:39
• @MichaelAlbanese what I actually ask for is an example. Mixed tensors are the ones that have both covariant and contravariant indices, according to Nolting's "Relativity" book. – physicist Dec 17 '18 at 20:41
• A mixed tensor would be, for example, the 4-dimensional representation matrix of the Lorentz transformation – physicist Dec 17 '18 at 20:43
• The example you give of a mixed tensor is not a mixed tensor, rather it has mixed signature (it is definite, but neither positive nor negative definite). – Michael Albanese Dec 18 '18 at 2:22
• Then what type of tensor is it, and could you provide me with an example of a mixed tensor? – physicist Dec 18 '18 at 3:23

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

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 $$T : \Gamma(T^*M)^p\times\Gamma(TM)^q \to C^{\infty}(M)$$. 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 $$\hat{\partial}_i = \frac{\partial}{\partial\hat{x}^i}$$, and $$\{d\hat{x}^1,\dots, d\hat{x}^n\}$$ is the dual basis of sections for $$T^*M|_U$$, so we get another collection of smooth functions $$\hat{T}^{i_1',\dots,i_p'}_{j_1',\dots,j_q'} := T(d\hat{x}^{i_1'},\dots, d\hat{x}^{i_p'}, \hat{\partial}_{j_1'},\dots, \hat{\partial}_{j_q'})$$ on $$U$$.

Note that $$\hat{\partial}_i = \dfrac{\partial x^k}{\partial \hat{x}^i}\partial_k$$ and $$d\hat{x}^j = \dfrac{\partial \hat{x}^j}{\partial x^k}dx^k$$ 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 $$\left(\dfrac{\partial\hat{x}}{\partial x}\right)^{-1} = \dfrac{\partial x}{\partial\hat{x}}$$, 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 $$r + s = p + q$$.