# Order of tensor indices

1. Feb 1, 2009

### emma83

Hello,

I don't understand what is the difference between. e.g. the (1,1)-tensor T_{a}^{b} and T^{b}_{a}, i.e. when the lower and upper indices are exactly the same but in another "vertical order", one slightly to the left and the other one slightly to the right.

2. Feb 1, 2009

### Fredrik

Staff Emeritus
The metric tensor is used to raise/lower indices, so

$$T_a{}^b=g_{ac}T^c{}_d g^{db}$$

where $g^{db}$ is defined by

$$g^{ab}g_{bc}=\delta^a_c$$

3. Feb 1, 2009

### emma83

Thank you, but this does not really answer my question, I fear:

In your first formula, the T_{a}^{b} on the left-hand side has the lower indice to the left and the upper to the right. This is a consequence of the fact that you "first" lower the "c" (which becomes an "a") and "then" upper the "d" (which becomes a "b") from the tensor on the right hand side. Is that correct ?
In other words, the "left-right" position actually results from the lowering-uppering operation, right ?

Now why, initially, do you have (for the T^{c}_{d} on the right-hand side) the "c" to the left and the "d" to the right ? This tensor is not a result of a lowering-uppering operation, so why these positions and not the opposite ones ? Would there be a difference with T_{d}^{c} (i.e. "d" on the left and "c" to the right ?)

(By the way, how do you type the nice formulae in this forum ?!)

Thanks a lot for your help!

4. Feb 1, 2009

### robphy

Use [ tex ] [ /tex ] tags (with no spaces).
Click on the following equation to see: $$T_a{}^b$$

5. Feb 1, 2009

### emma83

Thank you. So to make things clear, my question is:

What is the difference between $$T_{a}{}^{b}$$ and $$T^{b}{}_{a}$$ ?

6. Feb 1, 2009

### Fredrik

Staff Emeritus
(Edit: I got two equations wrong my first version of this post, and corrected them after emma83's reply below. I put red asterisks around those equations to show which ones they are).

$T_{a}{}^{b}$ is a component of a tensor T that acts on a pair $(v,\omega)$ where $v$ is a tangent vector and $\omega$ is a cotangent vector (a member of the dual space of the tangent space). We have *** $T_{a}{}^{b}=T(e_a,f^b)$ *** where $e_a$ is a basis vector of the tangent space, and $f^b$ is one of the dual basis vectors defined by $f^a(e_b)=\delta^a_b$.

$T^{b}{}_{a}$ is a component of a tensor T that acts on a pair $(\omega,v)$. We have *** $T^{b}{}_{a}=T(f^b,e_a)$ ***.

The notation gets confusing sometimes. You may be aware of "the abstract index notation", in which the first of these two tensors isn't written as T, but as $T_{a}{}^{b}$, just to make it clear from the notation what it's supposed to act on. The components are then written with greek indices, e.g. $T_\mu{}^\nu$, or at least that's the convention used in Wald's book.

Last edited: Feb 2, 2009
7. Feb 1, 2009

### emma83

Thanks a lot, I think it's getting clearer :-)
So if I understand well, the ordering in the indices of a tensor component reflects directly the ordering of the arguments in the argument list of the tensor ?
So for instance:

$$T^{b}{}_{a}{}^{cd}{}_{k}{}^{ig}=T(e_b,f^a,e_c,e_d,f^k,e_i,e_g)$$

Do you agree with that ?

8. Feb 2, 2009

### Fredrik

Staff Emeritus
Yes.

Oops, I see now that I messed up the part where I told you what the components of the tensor are. I can still edit my previous post, so I'll fix it there. The equation in your post should be

$$T^{b}{}_{a}{}^{cd}{}_{k}{}^{ig}=T(f^b,e_a,f^c,f^d,e_k,f^i,f^g)$$

Note that all indices appear at the same "altitude" (as superscripts or as subscripts), and in the same order, on both sides.

9. Feb 2, 2009

### emma83

Why that ? The indice for e.g. a contravariant vector $$e_{a}$$ itself (not its components $$e^{a}$$) is lower, right ?
Otherwise the summation notation doesn't work when a vector is represented in a basis, for instance as: $$U=U^{a}e_{a}$$.

So in this case the notation:
$$T^{b}{}_{a}{}^{cd}{}_{k}{}^{ig}=T(e_b,f^a,e_c,e_d, f^k,e_i,e_g)$$
is the correct one, since the arguments in the argument list on the right-hand side are the tensor objects (not their components).

Or am I wrong here ?

10. Feb 3, 2009

### Fredrik

Staff Emeritus
Yes, that's wrong. Consider e.g.

$$T(v,\omega)=T(v^\mu e_\mu,\omega_\nu f^\nu)=v^\mu \omega_\nu T(e_\mu,f^\nu)=v^\mu\omega_\nu T_\mu{}^\nu=T_\mu{}^\nu v^\mu\omega_\nu$$

That's the index free notation. (You put indices on tensor components and basis vectors/covectors, but not on the tensors themselves). Note that in the expression $v=v^\mu e_\mu$, $v^\mu$ is a component of a tangent vector, and $e_\mu$ is a tangent vector. (The components of $e_\mu$ are $\delta^\nu_\mu$, since $e_\mu=\delta^\nu_\mu e_\nu$). I've been trying to remember how the abstract index notation handles expressions like $$v=v^\mu e_\mu[/itex], but I can't think of a way that makes sense. I think the abstract index notation would just skip writing out the intermediate steps, and express the stuff above as [tex]T_a{}^b v^a\omega_b=T_\mu{}^\nu v^\mu\omega_\nu$$

(Perhaps someone who has a copy of Wald nearby, or in fresh memory, can confirm or deny this).

Last edited: Feb 3, 2009
11. Feb 3, 2009

### robphy

With the abstract-index notation, the indices are thought of as labels for slots in an argument list.

$$T_{a}{}^b{}_{cd}$$ is a multilinear map taking an ordered-tuple (vector,covector,vector,vector) [e.g. $$T_{a}{}^b{}_{cd}u^a\eta_b v^c w^d$$ ] into (say) the reals.
If you don't want to see any indices, one could write $$T(u,\eta,v,w)$$.

(There isn't a need to discuss basis vectors or components.)

See, for example, Ludvigsen: