# What is the trace of a second rank covariant tensor?

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1. Feb 9, 2016

### dwellexity

What is the trace of a second rank tensor covariant in both indices?
For a tensor covariant in one index and contravariant in another $T^i_j$, the trace is $T^k_k$ but what is the trace for $T_{ij}$ because $T_{kk}$ is not even a tensor?

2. Feb 9, 2016

### andrewkirk

Such a tensor is not readily interpreted as a linear operator on a single space, because if applied to a single (co)vector, it gives a result in the dual space of the input. There are various ways one could define something a bit like a trace, and call it a trace, for such a tensor. The definition would be an extension of the concept of a trace. The easiest way to do this is if there is a metric $g$ in the context of the question, in which case we could define the trace as $T^{ij}g_{ij}$. But whether that gives the desired result depends on the reason why one wants to extend the trace definition.

Is there a specific context for the question, in which a trace is needed, or is it just exploratory musing?

3. Feb 9, 2016

### dwellexity

Basically it is related to a homework problem I have. I need to write an arbitrary second rank covariant tensor as a sum of traceless symmetric tensor, an antisymmetric tensor and the trace. I wrote it as $[\frac{1}{2} (T_{ij} + T_{ji}) - \frac{1}{n} T \delta_{ij}] + [\frac{1}{n} T \delta_{ij}] + [\frac{1}{2} (T_{ij} - T_{ji})]$
T is the trace.

But then I thought how would the trace be represented in index notation?

Also, am I right to say that I can't really write the T as $g^{ij}T_{ij}$ because $T_{ij}$ is an arbitrary tensor and I don't know whether a metric is defined for it?

4. Feb 9, 2016

### andrewkirk

That helps. It seems to me that the lecturer is thinking of the tensors as just matrices, because the expression $\delta_{ij}$ has no meaning in a coordinate-free context.
In a matrix context, your solution appears to be correct.
You could point out to your lecturer that a linear map from one space to a different one is not a linear operator, and does not have a trace. But I doubt it would endear you to her, so better not do that.
This looks like one of those not infrequent questions where the more alert students have more problems than others, because they spot problems in the question that the others don't.
Your last sentence is correct too.

5. Feb 9, 2016