Index notation - I never know when to introduce a new symbol?

jeebs
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This isn't strictly a homework problem but anyway...
I'm reading through a QFT textbook that is using index notation, and sometimes a new index symbol will be introduced during some mathematics and it always throws me off. I'll give a simple example, take the Minkowski metric:

[tex]g^{\mu\nu} = \left(\begin{array}{cccc}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{array}\right)[/tex] and its inverse: [tex]g_{\mu\nu} = \left(\begin{array}{cccc}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{array}\right)[/tex]

We can multiply these 2 matrices together, ie. we could take [itex]g^{\mu\nu}g_{\mu\nu}[/itex] to get the identity matrix. However - and this confuses me - we could also take [itex]g^{\mu\nu}g_{\mu\nu}[/itex] to mean just the sum of the products of the matrix elements over both indices, as both are repeated:

[tex]g^{\mu\nu}g_{\mu\nu} = g^{00}g_{00} + g^{01}g_{01} + g^{02}g_{02} + g^{03}g_{03} + g^{10}g_{10} + g^{11}g_{11} + g^{12}g_{13} + g^{20}g_{20} + g^{21}g_{21} + g^{22}g_{22} + g^{23}g_{23} + g^{30}g_{30} + g^{31}g_{31} + g^{32}g_{32} + g^{33}g_{33}[/tex]

So, the first thing that confuses me is, how come we use indices when we refer to the full matrix [itex]g^{\mu\nu}[/itex], when normally we would just call a matrix (for example) [itex]A[/itex], and only mention indices [itex]i, j[/itex] when we want to refer to the [itex]i^{th}, j^{th}[/itex] element of the matrix, [itex]A^{ij}[/itex] ?
It seems to me that there is ambiguity here, when is [itex]g^{\mu\nu}g_{\mu\nu}[/itex] a matrix and when is it just a number?

Also, to get to the main part of my question, my book makes the statement that [itex]g^{\mu\nu}g_{\nu\rho} = \delta^{\nu}_{\rho}[/itex], the kronecker delta.
Here it has introduced a new index [itex]\rho[/itex]. I can see that this is true if I do the summation over [itex]\nu[/itex]:
[itex]g^{\mu\nu}g_{\nu\rho} = g^{\mu 0}g_{0\rho} + g^{\mu 1}g_{1\rho} + g^{\mu 2}g_{2\rho} + g^{\mu 3}g_{3\rho}[/itex]

then if we set, say, [itex]\mu = 0, \rho = 0[/itex], we get
[itex]g^{\mu\nu}g_{\nu\rho} = g^{0 0}g_{00} + g^{0 1}g_{10} + g^{0 2}g_{20} + g^{0 3}g_{30} = (1)(1) + (0)(0) + (0)(0) + (0)(0) = 1[/itex]

or if we set, say, [itex]\mu = 0, \rho = 1[/itex], we get
[itex]g^{\mu\nu}g_{\nu\rho} = g^{0 0}g_{01} + g^{0 1}g_{11} + g^{0 2}g_{21} + g^{0 3}g_{31} = (1)(0) + (0)(-1) + (0)(0) + (0)(0) = 0[/itex]

So clearly the Kronecker delta condition is satisfied, so the statement [itex]g^{\mu\nu}g_{\nu\rho} = \delta^{\nu}_{\rho}[/itex] is true. However, if I was writing out my own solution to a problem that involved index notation, I would never know to introduce a new index symbol myself. It's just lucky that the textbook told me and I could verify it with an explicit calculation.

Can anyone explain to me how to know when a new index symbol should be introduced?
 
Last edited:
on Phys.org
Thanks in advance! A:The idea of introducing a new index is to make the statement of the problem more concise. The simple example you have provided is actually just a special case of a much more general idea.In general, when dealing with tensors (which are objects that can be indexed by multiple indices) it is often useful to introduce a new index in order to avoid having to write out the same expression multiple times. For example, if we have a four-dimensional tensor $T_{\mu \nu \rho \sigma}$ and want to contract it with its inverse $T^{\mu \nu \rho \sigma}$ we could write out the full expression as$$T_{\mu \nu \rho \sigma} T^{\mu \nu \rho \sigma} = T_{0000} T^{0000} + T_{0001} T^{0001} + \dots + T_{3333} T^{3333}.$$This looks unwieldy, so it is often more convenient to introduce a new index $\lambda$ and rewrite the expression as$$T_{\mu \nu \rho \sigma} T^{\mu \nu \rho \lambda} = T_{0000 \lambda} + T_{0001 \lambda} + \dots + T_{3333 \lambda}.$$The idea here is that we are saying that the expression $T_{\mu \nu \rho \sigma} T^{\mu \nu \rho \lambda}$ should be evaluated for all possible values of $\lambda$, and then the result should be summed over all possible values of $\lambda$. This is equivalent to the original expression, but it is much more concise.
 

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