# Is there any reason for tensor indices being ordered one way or the other?

if I have an expression

gca * dwab

is there any reason for the resultant tensor should be dwcb vs. dwbc?

diazona
Homework Helper
Yes, because not all tensors are symmetric. With a non-symmetric tensor, Tcb ≠ Tbc, so if you change Tcb to Tbc, you've changed the value.

I guess the rule would be to keep left indices as left indices and right indices as right indices. In the example you give, I look at the sequence of indices "caab" and think of the two a's in the middle as "canceling out" - although that doesn't always work, when you're dealing with more complicated tensor products.

Think what happens when you transpose a matrix. If the matrix is symmetric about the main diagonal, its components are unchanged, otherwise there will be a difference. If you know that the tensor you're dealing with is symmetric in a given pair of indices, you're free to swap them; if not, just remember to keep them in the same horizontal order. If it's antisymmetric in a given pair of indices, swapping them has the same effect as multiplying by -1, so you can swap them and multiply by -1 to get back to where you started.

thankyou,

my question is more about why the order matters and how to determine the proper order in general. If you look at the first expression, its an expression to lower an index, but when this ndex gets lowered which side would it be on?

If you look at the first expression, its an expression to lower an index, but when this ndex gets lowered which side would it be on?

gca dwa b = dwcb

In general, the horizontal order doesn't change. Just turn a into c and bring it straight down.

what if I had written it

in general when the indices come down do they go to the front of the line? or is there a situation where they would go to the back of the line?

and what about the covariant indices, if they are raised do they go to the front or back of the line at the top?

I think the above two questions are really where I'm getting hung up...

thankyou

George Jones
Staff Emeritus
Gold Member
in general when the indices come down do they go to the front of the line? or is there a situation where they would go to the back of the line?

and what about the covariant indices, if they are raised do they go to the front or back of the line at the top?

The indices move straight vertically.

thankyou,

I still dont see the difference between a rank (1,1) tensor such as wab and wba, I may have to do more reading.

thankyou,

I still dont see the difference between a rank (1,1) tensor such as wab and wba, I may have to do more reading.

Saying that a tensor is of type $$(r,s)$$ means that it has $$r$$ number of contravariant indices and $$s$$ number of covariant indices.
Contravariant and covariant vectors transform slightly different under coordinate transformations.

A contravariant vector $$\mathbf{v}=v^{\alpha}$$ transforms according to
$$v^{\alpha}=x^{\alpha}_{, \beta}v^{\beta}$$
while a covariant vector $$\mathbf{u}=u_{\alpha}$$ transforms as
$$u_{\alpha}=y^{\beta}_{, \alpha}u_{\beta}$$

It has to do with which vector space they live.
For the vector space $$V in a fixed basis [tex] \{\epsilon_{\alpha}\}$$, contravariant vectors are the row vectors $$x^{\alpha}$$ of contravariant components.

The dual space $$V^{*}$$ will have a dual base $$\{ e^{\alpha} \}$$ where the the covariant vectors are given by
$$\mathbf{v}=\{e^{\alpha} \}v_{\alpha}$$

If $$V$$ and $$V^{*}$$ are isomorphic, then the space of tensors of a finite rank are symmetric about their indices....the tensors are invariant when you permute the indices.

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If $$V$$ and $$V^{*}$$ are isomorphic, then the space of tensors of a finite rank are symmetric about their indices....the tensors are invariant when you permute the indices.

That seems to go against what I learnt. Here's how I currently understand it. All corrections welcome.

In general, $T_{\mu \nu} \neq T_{\nu \mu}$, right? There will always be an isomorphism between $V$ and $V^{*}$ since they're over the same field and of the same dimension, therefore, if $\mathbb{K}$ is the field, and $n$ the dimension, $V [/tex] and [itex] V^{*}$ are each isomorphic to $\mathbb{K}^n$, which implies that $V$ is isomorphic to $V^{*}$. If we have a metric, $g$, there will be a natural (canonical) isomorphism defined by the metric and its inverse:

$$g:V\to V^*: g_{\mu \nu} v^\nu=v_\mu$$

$$g:V^* \to V : g^{\mu \nu}v_\nu=v^\mu$$

If no metric is defined, only the horizontal order of indices relative to others at the same height matters:

$$T^{\alpha \beta}\;_{\gamma}=T_{\gamma}\;^{\alpha \beta}=T^{\alpha}\;_{\gamma}\;^{\beta}$$

so you don't need to keep track of them and could just write $T^{\alpha \beta}_{\gamma}[/tex]. This is because there's no natural answer to the question: "Which tensor with an up index here corresponds to the tensor with a down index?" But if a metric is defined, that question does have a definite answer, so you do have to keep track of the horizontal order of indices relative to those of both heights, up and down, because $$g_{\alpha \mu} T^{\alpha}\,_{\nu} = T_{\mu \nu}$$ $$g_{\alpha \mu} T_{\nu}\,^{\alpha} = T_{\nu \mu}$$ and $$T_{\mu \nu} \neq T_{\nu \mu}$$ unless it [itex]T$ happens to be symmetric. But in general, there's no guarantee that $T$ will be symmetric.

jcsd
Gold Member
Actually there isn't always an isomorphism between V and V*!

But that's more of an interesting factoid as we're clearly only interestred in the finite dimensional case where they are always isomorphic.

Actually there isn't always an isomorphism between V and V*!

But that's more of an interesting factoid as we're clearly only interestred in the finite dimensional case where they are always isomorphic.

Ooh, thanks for that proviso, I hadn't thought of whether it would be different with infinite dimensions. Most of my reading has been confined to the case of finite dimensions. Is the rest okay?

Just by way of confirmation: "The index staggering is needed for when a metric is introduced since spaces are needed for the raising and lowering of indices" (Roger Penrose: The Road to Reality (Vintage 2005), note 14.3, p. 323).