Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose we have some tensor with components $T^{ij}$. Then suppose that we also have $T_{ij}$.

When would $T^{ij}T_{ij} = (T^{ij})^2 = (T_{ij})^2$?

share|cite|improve this question
up vote 4 down vote accepted

Note; I'll use the summation convention throughout here.

In the context of differential geometry, the indices on tensorial objects are raised and lowered with the metric on the space (manifold) being studied. So for example $$ T^i_{\phantom i j} = g^{ik}T_{kj} $$ and $$ T^{ij} = g^{ik}g^{jl}T_{kl} $$ Notice that if the metric is simply that of Euclidean space, namely if $g_{ij} = \delta_{ij}$, then raising and lowering does not change the numerical values of tensor components. In particular, one would have $$ T^{ij} = T_{ij} $$ Notice that in expressions like $$ T^{ij}T_{ij} $$ both indices are being summed over, where as in the expressions $$ (T^{ij})^2, \qquad (T_{ij})^2 $$ one usually (this is actually a matter of notational preference) doesn't intend for their to be any implied summation, so typically its notationally safe to assume that $$ T^{ij}T_{ij}\neq (T^{ij})^2, \qquad T^{ij}T_{ij}\neq (T_{ij})^2 $$ but if the metric satisfies $g_{ij} = \delta_{ij}$, then it is true that $$ (T^{ij})^2=(T_{ij})^2 $$

share|cite|improve this answer
In general, I would understand summation in $(T_{ij})^2$. In gauge theory, for example, it's almost standard to write $S = \int -\frac{1}{4} (F_{\mu\nu})^2.$ What would it be, otherwise? – Vibert Mar 12 '13 at 21:20
I guess it varies by field. (Personally, I'm also used to seeing $(\cdot)^2$ used to indicate a trace over all written indices, $(T_{ij})^2 = T_{ij}T^{ij}$.) – David Z Mar 12 '13 at 21:22
slight correction to your last line--if the metric is Euclidean <b>and the coordinates are Cartesian</b>, then it is true that... – Jerry Schirmer Mar 12 '13 at 21:28
@Vibert Yeah I've seen that notation used quite a bit as well; it's simply a matter of notational taste it seems. – joshphysics Mar 12 '13 at 22:24
@JerrySchirmer Thanks; I just changed it to if $g_{ij} = \delta_{ij}$ so as to avoid confusion. – joshphysics Mar 12 '13 at 22:25

i,j run only over space indexes and we are working in realms of special relativity then all of them will have same signature take metric to be g=diag(1(time),-1(x),-1(y),-1(z)). Hence does not make a diff if you write it in top or bottom. And in fact (i,j) are latin symbols which when used suggest space indexes. And ($\mu,\nu$) i.e. greek indices when written suggest space and time indexes

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.