Im taking GR for the first time and it is definitely throwing me for a loop. The question I am working on is this:

Prove that if a contravariant tensor $A^{uv}$ is symmetric, then it remains symmetric under Lorentz transformations. Is this also true for a mixed tensor $A_{u}{}^{v}$?

I'm working on the first part. I have a tensor: $$A^{uv} = A^{vu}$$

And I want to show that it is symmetric under a Lorentz transformation. For reasons which I do not know, I need to multiply by 2 lorentz transforms and show that the outcome is still symmetric. I am very bad with tensor notation and as such I am sure what I am going to write next will make experts in the field cringe. Is this the correct way to multiply $A^{uv}$ by 2 lorentz transforms?

$$A'^{uv} = {{a_\alpha}^\beta}{{a_\rho}^\sigma}A^{uv}$$

I know this is wrong, however I'm not sure why. Here is my thought process: The two a's correspond to the 2 different lorentz transforms, and they are "operating" on the original tensor A to give the new, transformed tensor. The reason why I know this is wrong is because there is no "cancellation" of any of the sub/superscripts, however, I'm not sure how to fix it.

I'll try to be brief, but while I am at it, what does it mean for a tensor (in this case a matrix) to be contravariant? I was struggling with the difference between contravariant and covariant vectors, and someone much more knowledgeable explained it to me in this way: "A contravariant vector is a covariant vector which has been operated on by the metric you're working in. Normally we don't think about it because the euclidean metric is the identity matrix, however in SR the metric commonly used is the minkowski metric". What would be the analogy with a matrix?

  • $\begingroup$ They did teach you how to implicitly sum over the same index, didn't they? Did you learn how this is equivalent to matrix and vector products? If you haven't, then you are trying to read "Moby Dick" without having learned you ABCs... $\endgroup$
    – CuriousOne
    Feb 20, 2016 at 1:11
  • $\begingroup$ Yes, I know that a repeated index is a sum over that index, if that is what you are asking. $\endgroup$
    – J. Doe
    Feb 20, 2016 at 1:13
  • $\begingroup$ So then use what you have learned. Your second expression simply builds a tensor with six indices. You need to have something to contract over and the number of upper and lower indices on both sides has to agree. $\endgroup$
    – CuriousOne
    Feb 20, 2016 at 1:14
  • $\begingroup$ $$A'^{uv} = {{a_\alpha}^{u}}{{a_\sigma}^{v}}A^{\alpha \sigma}$$ $\endgroup$
    – J. Doe
    Feb 20, 2016 at 1:17
  • $\begingroup$ That looks a lot better now. $\endgroup$
    – CuriousOne
    Feb 20, 2016 at 1:17

3 Answers 3


Imagine you have a space. On it you could have a scalar field. And from that scalar field you could imagine taking its gradient.

In your elementary math classes they might have said that a gradient is a vector field. But you probably noticed that the formulas could look really weird in different coordinate systems, so lets step up and ask what you do with a gradient.

When you have a gradient what it does is take a scalar field and gives you a vector field, specifically at each point it tells you the rate the original scalar field changes in the direction of the vector field.

And locally that means that there are things like gradients (except local) called covectors and things like vectors. And they combine together to give scalars. Much like how column vectors (nx1 matrices) and row vectors (1xn matrices) combine to give you a scalar (1x1 matrix).

And none of this depended on a coordinate system or a basis.

So now what if you pick a basis. Each of those covectors or vectors could be written in terms of basis vector or basis covectors. But the action of taking a covector and a vector and getting a scalar shouldn't depend on a basis or a coordinate system.

So if you could change the coordinates of a vector by multiplying by a matrix on the right then you should change the covectors by multiplying by the inverse of that matrix on the right.

So now when you have a tensor you do the same math as multiplying by the matrix or its inverse. You have some scalars you multiply and you add up.

So if you are used to thinking of a covector as a function that takes vectors and gives scalars. And thinking of a vector as a function that takes covectors and gives scalars. Then you can generalize to a function that takes a vector and gives you a covectors (i.e. it takes two vectors in a bilinear fashion and gives a scalar).

And these transform in the particular way they have to transform. And how is that. Basically by doing matrix multiplication on the right by that inverse matrix. Why. Because it is the way to have the function give the same value on the same vectors. The vectors get multiplied on the left by a matrix, so the covectors have to get multiplied by the inverse matrix on the right. Because the goal is to have the same vectors give the same scalars.

  • $\begingroup$ I'm not sure this is an answer, but it's better than an answer because it actually tries to explain what's happening rather than the usual physicist's approach of focussing on the index gymnastics. $\endgroup$
    – user107153
    Feb 20, 2016 at 11:27

Have a look at Timaeus's answer for an insight into what is actually happening, but I'll just focus on the index gymnastics.

So you have the components $A^{\mu\nu}$ of a tensor in some basis, and you know that $A^{\mu\nu} = A^{\nu\mu}$: it's symmetric.

Well, you want the components $A'^{\alpha\beta}$ in some other basis. What do you know about the transformation between the two bases? Well, you know that the basis vectors $\left\{\vec{e}_\mu\right\}$ transform by some nonsingular matrix $\Lambda$: $\vec{e}'_\alpha = {\Lambda_\alpha}^\mu\vec{e}_\mu$. And it follows from this that the components of a $\binom{2}{0}$ tensor transform like this:

$$ {{T'}^\alpha}^\beta = T^{\mu\nu} {\left(\Lambda^{-1}\right)_\mu}^\alpha {\left(\Lambda^{-1}\right)_\nu}^\beta $$

So, let's just call ${\left(\Lambda^{-1}\right)_\mu}^\alpha = {a_\mu}^\alpha$, to agree with your notation.

Now you can use the above formula to work out what $A'^{\alpha\beta}$ is, and also what $A'^{\beta\alpha}$ is, in terms of $A^{\alpha\beta}$ and $A^{\beta\alpha}$, and you will see immediately that if $A^{\alpha\beta} = A^{\beta\alpha}$ then the same is true for the components in the new basis: basis transformations preserve this symmetry.

You might want to look at this which deals with how to go from the geometrical picture to the index gymnastics in more detail. Disclaimer: I wrote it.

  • 1
    $\begingroup$ I would like to point out that it is probably easiest to see this with trying to transform tensor components of the "zero" tensor field. Let us assume $A^{ij}$ is symmetric, then $0=A^{ij}-A^{ji}$, now if we transform the $A$s, we get $A^{kl}a_k^i a_l^j-A^{lk}a_l^j a_k^i=(A^{kl}-A^{lk})a_l^j a_k^i=0$, which is true because the quantity in the brackets is known to be zero, and so if $A'^{ij}=A^{kl}a_k^i a_l^j$, we have $A'^{ij}-A'^{ji}=0$, ergo the transformed quantities are also symmetric. $\endgroup$ Feb 20, 2016 at 13:35

I think mainly you don't understand what is a tensor. A tensor is not an object with $n-$upper and $m-$down indices, like a matrix or things like that. So, what is a tensor? Well, first we need a vector space with some inner product, $(\cdot,\cdot)$, $\mathbb R^3$ will work just fine. Because this is an inner vector space, we can construct its dual vector space $\mathbb R^{*\,3}$. Take an orthonormal basis $\{\textbf{e}_1,\,\textbf{e}_2,\,\textbf{e}_3\}$ for $\mathbb R^3$, so any vector of this space could look like this $\textbf{v}=v^1\textbf{e}_1+v^2\textbf{e}_2+v^3\textbf{e}_3$, where $v^i$ are just numbers. Now, a basis $\{\theta^1,\,\theta^2,\,\theta^3\}$ of its dual vector space $\mathbb R^{*\,3}$ is a set of linear maps that satisfy $\theta^i(\textbf{e}_j)=\delta^i_j$. Now, this last condition need to be independent of any change of basis. So, if you have any other basis $\{\textbf{e}'_1,\,\textbf{e}'_2,\,\textbf{e}'_3\}$ it relates to the first by $\textbf{e}'_i=\sum_{k=1}^3\, a_i^{\hspace{.15cm}k}\,\textbf{e}_k$, where $a_i^{\hspace{.15cm}k}$ are the COMPONENTS of the matrix of transformation, and then $\{\theta'^1,\,\theta'^2,\,\theta'^3\}$ need to be related to the original basis by $\theta'^j=\sum_{k=1}^3\, (a^{-1})^j_{\hspace{.15cm}k}\,\theta^k$, where $(a^{-1})^j_{\hspace{.15cm}k}$ are the COMPONENTS of the inverse matrix of transformation. Only then, $\theta'^i(\textbf{e}'_j)=\sum_{m=1}^3\sum_{n=1}^3(a^{-1})^i_{\hspace{.15cm}n}a_j^{\hspace{.15cm}m}\,\theta^n(\textbf{e}_m)=$ $\sum_{m=1}^3\sum_{n=1}^3(a^{-1})^i_{\hspace{.15cm}n}a_j^{\hspace{.15cm}m}\,\delta^n_m$ $=\delta^i_j$. Check carefully the indices in the last expressions!

Physicists loooove easy notation, so in the next part I'm going to use the Einstein's index convention.

Now, I will talk about what is a 2-contravariant tensor $T$ (yes, without index), respect to the vector space I'm working. This is a multilinear map from $\mathbb R^{*\,3}\times\mathbb R^{*\,3}$ to $\mathbb R$. Now, this object (and any other kind of tensor) doesn't care about the basis you choose to work with, at the end a tensor is just a number. Taking $\omega, \,\alpha\in\mathbb R^{*\,3}$, then $\omega=w_i\,\theta^i$ and $\beta=b_j\,\theta^j$. So


It is costumary to define $T^{ij}=T(\theta^i,\,\theta^j)$ and this is what you maybe know as a tensor (which is wrong). This are known as the COMPONENTS of the tensor $T$. As you can see, this aren't invariant under a change of basis and the same goes for $w_i$ and $b_k$, but $T(\omega,\,\beta)$ as a whole it is (this is the beauty of tensors). So, under a change of basis $w'_i=a_i^{\hspace{.15cm}n}\,w_n$, $b'_i=a_i^{\hspace{.15cm}m}\,b_m$, and

$T'\,^{ij}=T(\theta'^i,\,\theta'^j)=T((a^{-1})^i_{\hspace{.15cm}n}\,\theta^n,\,(a^{-1})^j_{\hspace{.15cm}m}\,\theta^m)=$ $(a^{-1})^i_{\hspace{.15cm}n}(a^{-1})^j_{\hspace{.15cm}m}T(\theta^n,\,\theta^m)=$ $(a^{-1})^i_{\hspace{.15cm}n}(a^{-1})^j_{\hspace{.15cm}m}T^{mn}$.

  • $\begingroup$ You don't, in fact, need the vector space to have an inner product to define the dual space: the dual space is just the space of linear functions from elements of the original space to the reals (or whatever the scalars the space is defined over are). $\endgroup$
    – user107153
    Feb 20, 2016 at 15:18
  • $\begingroup$ yes, but he will need it to define a Lorentz transformation... $\endgroup$
    – raul
    Feb 20, 2016 at 15:20
  • $\begingroup$ However it's easy to show (see my answer) that any basis change preserves symmetry: the result is much more general. $\endgroup$
    – user107153
    Feb 20, 2016 at 17:19
  • $\begingroup$ I'm not trying to prove the symmetry here. There are several answers already about that. The way I see it, the main problem of the OP is that he don't know anything about tensors. So, how can he understand any answer? $\endgroup$
    – raul
    Feb 22, 2016 at 18:19

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