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This is from my notes, which I don't fully understand:

  1. It is straightforward to check that (anti)symmetry is a coordinate-independent notion, e.g., if the components of a tensor are symmetric in some coordinate system, they are symmetric in all.

  2. note that it only makes sense to discuss symmetry of pairs of contra- or co-variant indices, but not a mix of the two.

I'm guessing in 1 that the notes is referring to symmetry in all coordinates, not just two (or any subset) specific coordinates. I'm happy that this is the case for the symmetric case, but could someone explain this more in depth in the anti-symmetric case?

I'm not sure why 2 is the case too, surely we can construct a tensor which is symmetric over a pair of contra and co variant indices? ie $A^\mu_{\lambda} = A^\lambda_\mu$

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I think you are confused by the meaning of the first statement. It does not say that (anti-)symmetry in two indices implies (anti-)symmetry in all indices; one is perfectly free to have tensors that are (anti-)symmetric in any number of indices, as long as these are of the same type. Instead, it refers to the fact that symmetries of tensors are untouched by coordinate transformations.

A (p,q)-tensor is a linear map from p covectors and q vectors to (in a GR context) the real numbers. For example, let's look at a (1,2)-tensor $A^{\mu}_{\ \ \ \nu \rho}$. We can contract this object with a covector $W_{\mu}$ and two vectors $U^{\nu}$ and $V^{\rho}$ to form the real number $A^{\mu}_{\ \ \ \nu \rho} W_{\mu} U^{\nu} V^{\rho}$. $A$ being (anti-)symmetric in its lower indices means that switching $U$ and $V$ leaves the outcome unchanged, apart from a sign in the antisymmetric case: $A^{\mu}_{\ \ \ \nu \rho} W_{\mu} U^{\nu} V^{\rho} = \pm A^{\mu}_{\ \ \ \nu \rho} W_{\mu} V^{\nu} U^{\rho}$. This corresponds to the requirement that $A^{\mu}_{\ \ \ \nu \rho} = \pm A^{\mu}_{\ \ \ \rho \nu}$.

Since (co)vector components are coordinate-dependent, the tensor components should be as well. In general, a tensor transforms by contraction with the coordinate changes $\partial x ^{\mu} / \partial y ^{\sigma}$. The notion of a tensor being (anti-)symmetric could well be coordinate-dependent: a tensor that is symmetric in some coordinate system could lose its symmetry in some other coordinate system. The content of your first statement is that this is not the case. This is rather easily shown.

Let's focus on a (0,2)-tensor $A_{\mu \nu}$ for now. Under a change of coordinates $\partial x ^{\mu} / \partial y ^{\sigma}$ this tensor becomes: $$ A_{\rho \sigma } = \frac{\partial x^{\mu}}{\partial y ^{\rho}} \frac{\partial x^{\nu}}{\partial y ^{\sigma}} A_{\mu \nu}. $$ Now, if $A_{\mu \nu}$ is symmetric, that is, $A_{\mu \nu} = A_{\nu \mu}$, $$ A_{\rho \sigma } = \frac{\partial x^{\mu}}{\partial y ^{\rho}} \frac{\partial x^{\nu}}{\partial y ^{\sigma}} A_{\mu \nu} = \frac{\partial x^{\mu}}{\partial y ^{\rho}} \frac{\partial x^{\nu}}{\partial y ^{\sigma}} A_{\nu \mu}. $$ We can now switch the $\mu$ and $\nu$ indices, since these are dummy indices which we sum over, to obtain $$ A_{\rho \sigma } = \frac{\partial x^{\nu}}{\partial y ^{\rho}} \frac{\partial x^{\mu}}{\partial y ^{\sigma}} A_{\mu \nu} = A_{\sigma \rho }, $$ showing that $A$ is also symmetric in the new coordinates.

If $B_{\mu\nu}$ is an antisymmetric (0,2)-tensor, we would get a minus sign; otherwise, the procedure to show antisymmetry of $B_{\rho \sigma}$ is identical.

Your second statement means that it does not make sense to compare upper with lower indices, since this would correspond to comparing vectors with covectors. Sure, you can construct a list of numbers that (numerically) satisfies $A_{\mu}^{\nu} = A_{\nu}^{\mu}$, but this 'symmetry' would not have any sensible meaning, since the index types of the LHS and RHS do not match.

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It's useful to remember that a $(p,q)-$tensor can be thought of as a multilinear map which eats $p$ covectors and $q$ vectors and spits out a real number: $$ T : \underbrace{V^* \times V^* \times \ldots \times V^*}_{p\text{ copies}} \times \underbrace{V \times V \times \ldots \times V}_{q\text{ copies}} \rightarrow \mathbb{R}$$

For example, a (0,2)-tensor eats two vectors and spits out a real number. A standard example of such a tensor is the metric $\bf{g}$. Because it is a multilinear map, we have that

$${\bf{g(X,Y)}}={\bf{g}}(X^i\hat e_i,Y^j \hat e_j) =X^i Y^j {\bf{g}}(\hat e_i,\hat e_j) \equiv X^i Y^j g_{ij}$$

where we define $g_{ij}$ to be what we get when we take the tensor ${\bf{g}}$ and plug the basis vector $\hat e_i$ into the first slot and the basis vector $\hat e_j$ into the second slot.

Similarly, consider the Riemann curvature tensor, which eats one covector $\boldsymbol \omega$ and three vectors $\bf{X,Y,Z}$ (making it of type (1,3)).

$${\bf{R}(\boldsymbol{\omega},X,Y,Z)} = {\bf{R}}(\omega_i\hat \epsilon^i,X^j\hat e_j, Y^k\hat e_k, Z^l \hat e_l) = \omega_i X^jY^kZ^l \cdot {\bf{R}}(\hat \epsilon^i,\hat e_j, \hat e_k, \hat e_l) \equiv \omega_i X^jY^kZ^l \cdot R^i_{\ \ jkl}$$


From this perspective, the answer to your question becomes clear. When we say that $R^{i}_{\ \ jkl}$ is antisymmetric in its last two indices, what we mean is that if we swap the basis vectors which we plug in to its last two slots, then the results differ by a minus sign.

It also means that saying that it's symmetric (or antisymmetric) in the first two indices doesn't make any sense, because the first slot is for covectors and the second slot is for vectors. You can't plug a vector into the first slot, and you can't plug a covector in the second slot, so the notion of swapping doesn't work.

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    $\begingroup$ I agree with your answer. A coordinate-free approach (among other things) often spares the need to fiddle with indices and with trasformation properties. It also makes some things obvious, like meaninglessness of symmetry between an upper and a lower index. Just a minor point: $R^i{}_{jkl}$ is antisymmetric in its last indices, not symmetric. $\endgroup$ – Elio Fabri Oct 17 '18 at 10:20
  • $\begingroup$ @ElioFabri Whoops, my mistake. Thanks for the correction. $\endgroup$ – J. Murray Oct 17 '18 at 10:49
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I prefer to use this notation. The transformation of a Tensor components is:

$S=R\,T\,R^T \quad\quad(1)$

R is the transformation matrix

T is the Tensor .

If $S=S^T$ then S is symmetric tensor.

If $S=-S^T$ then S is anti symmetric tensor.

$S^T=\left(R\,T\,R^T\right)^T=R\,T^T\,R^T \quad\quad(2)$

But if T is symmetric tensor then $T=T^T$, then follows (equations (1) and (2)) that $S=S^T$ is also symmetric tensor.

If T is anti symmetric tensor then $T=-T^T$, then follows that $S=-S^T$ is also anti symmetric tensor

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  • $\begingroup$ This does not hold literally for a general tensor, but only for a (0,2)-tensor. $\endgroup$ – Stijn B. Oct 16 '18 at 23:00

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