Something like the trace of a tensor

Question

I'm going through some Lorentzian geometry and had some questions about different ways of getting a scalar out of a $(0, 2)$-tensor. Let us denote it by $T_{ab}$. Suppose we're on a pseudo-Riemannian manifold $(M, g)$ of dimension $n + 1$. The most (?) natural way of getting a scalar out of $T$ is the trace: $$ \text{tr}T = g^{ab}T_{ab}. $$ In GR I often see a different scalar appearing, namely $$ T_{ab}T^{ab}. $$ Is there a name for this operation? Is it just contraction of $T$ with its dual? Is there some way of understanding it as a norm, or anything like that?

Second: Is there a relationship between this scalar and $\text{tr}T$?

Finally, what is meant by $|T|$? I see this appearing in some papers.

Any help is much appreciated!

Answer 1

A Riemannian metric is, among other things, an inner product on $(1,0)$ tensors (i.e. tangent vectors). $$ \langle u,v\rangle=g_{ab}u^av^a\ \ \ \ \ \ \ u,v\in T^1_0M $$ This generalizes naturally to an inner product on tensors of any rank, given by $$ \langle U,V\rangle=g_{a_1c_1}\dots g_{a_kc_k}g^{b_1d_1}\dots g^{b_ld_l}U^{a_1\dots a_k}{}_{b_1\dots b_l}V^{c_1\dots c_k}{}_{d_1\dots d_l} $$ $$ U,V\in T^k_l M $$ The expression you write down is just $\langle T,T\rangle$, which we may also write as $\|T\|^2$ where $\|\ \ \|$ is the norm induced by the inner product.

The symbol $|T|$ may denote the above norm, but not always. For example, it's common to write the Riemannian volume form as $$ dV_g=\sqrt{|g|}\ dx^1\wedge\dots\wedge dx^n $$ Here $|g|$ denotes the absolute value of the determinant of the coordinate representation of $g$. This "determinant" is coordinate dependent, and is frequently used when writing integral expressions in coordinates.

Answer 2

We can look at this question in a slightly more general situation, where on a manifold $M$ of dimension $d$ we are given a non-degenerate symmetric $(0,2)$-tensor $g_{ab}$, and a $(0,2)$-tensor $T_{ab}$. Here I use the so-called abstract index notation. The non-degeneracy of $g_{ab}$ means that there is a $g^{ab}$ such that $g_{ac}g^{cb} = \delta_a{}^b$ (the Kronecker symbol, representing the identity map $\delta_a{}^b \colon V^a \to V^b$ where for the moment let $V^a$ be the tangent bundle of $M$).

So, what can we potentially make out of tensor $T$ using only the given structure $g$?

Naturally, we would like to keep things within certain computability constraints and decide to restrict our freedom to all sorts of finite linear combinations of partial contractions which have the following form:

$$ g_{ab} \dots g_{cd} g^{ef} \dots g^{kl} T_{rs} \dots T_{pq} $$

where some of the upper indices may coincide with some lower indices whenever it makes sense.

Some examples of such contractions would be the corresponding linear transformation $T \colon V \to V$ given by $T_a{}^b = g^{bc} T_{ac}$, its trace $\mathrm{T} := T_a{}^a$, which is the same as $\mathrm{tr_g}T = g^{ab}T_{ab}$, the square of $T$ given by $(T^2)_a{}^b = T_a{}^c T_c{}^b$, and so on.

In particular, we obtain a sequence $T^k$ of powers of $T$:

$$ (T^0)_a{}^b := \delta_a{}^b, (T^{k+1})_a{}^b := (T^k)_a{}^cT_c{}^b,k \ge 0 $$

Notice, that all these considerations are essentially the multi-linear algebra. We proliferate some quantities from tensor spaces over $M$, which are known to be not irreducible. I will not go into full details here (my goal is to give you some hints), but let me just mention that

$$ t_{ab} = t_{[ab]} + t_{(ab)} = t_{[ab]} + t_{(ab)_{0}} + \tfrac{\mathrm{tr}t}{d}g_{ab} $$

for any tensor $t_{ab}$. Here $t_{(ab)} := \tfrac{1}{2}(t_{ab} + t_{ba})$ and $t_{[ab]} := \tfrac{1}{2}(t_{ab} - t_{ba})$ are the symmetric and the anti-symmetric parts of $t_{ab}$, and $t_{(ab)_{0}} := t_{(ab)}-\tfrac{\mathrm{tr}t}{d}g_{ab}$ is the trace-free part of $t_{(ab)}$.

Furthermore, only a finite number of the powers $T^k$ are linearly independent. This fact is known as the Cayley-Hamilton theorem (in disguise). In order to see this in a slick way, one exploits the fact that in dimension $d$ any anti-symmetric $(0,d+1)$-tensor vanishes identically. Thus, we can look at

$$ 0 = T_{[{a_1}}{}^{a_1} \dots T_{{a_d}}{}^{a_d} \delta_{{a_{d+1}}]}{}^{a_{d+1}} $$

and expand it to see the terms.

Personally, I would call $T_{ab} T^{ab}$ the trace of the square of $T$ and denote it as $\mathrm{tr} T^2$ or, more rigorously, as $\mathrm{tr_g} (T^2)$. Hence, the length (or norm) of $T$ is given by the formula $|T| = \sqrt{\mathrm{tr_g} (T^2)}$.

There are relationships between $\mathrm{tr_g} (T)$, $\mathrm{tr_g} (T^2)$, and other scalars, such as $\det{T}$. These relationships come from the Cayley-Hamilton identity, and therefore, dimensionally-dependent.

You can find more examples and explanations in my other answers, e.g. here, where I mention a relationship between the quantities in question for the case of dimension $2$:

$$ (\mathrm{tr}T)^2 - \mathrm{tr}(T^2) = 2\,\det T $$