# What is the canonical form of the metric?

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I am reading Spacetime and Geometry : An Introduction to General Relativity – by Sean M Carroll and he writes:
Quote: A useful characterisation of the metric is obtained by putting ##g_{\mu\nu}## into its canonical form. In this form the metric components become $$g_{\mu\nu} = \rm{diag} (-1, -1,...-1,+1,+1, ... +1,0,0, ... ,0)$$where "diag" means a diagonal matrix with the given elements. End quote.

Wikipedia tells me to "Write ## \rm{diag} (a_1, ..., a_n)## for a diagonal matrix whose diagonal entries starting in the upper left corner are ##a_1, ..., a_n##." So Carroll's expression seems to imply a diagonal matrix with with a minimum size 9x9 and one extra row and column wherever one of his .'s takes a value. Or something like $$\begin{pmatrix} -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & . & 0 \\ 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & . & 0 \\ . & . & . \\ 0 & 0 & 0 & 0 & -1& 0 & 0 & 0 & . & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & . & 0 \\ . & . & . \end{pmatrix}$$ In general Carroll does not assume that ##\mu,\nu## have a range such as 1,2,3 or 0,1,2,3 until he gets to examples which are much simpler.

I don't think my interpretation is correct. Can anybody cast any light for me?

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He means exactly the same thing as the wiki article.

Nugatory
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So Carroll's expression seems to imply a diagonal matrix with with a minimum size 9x9 and one extra row and column wherever one of his .'s takes a value.
No, this is not what is intended. Each set of numbers (-1,1,0) can correspond to any quantity of that number. For a non-degenerate metric there would be no zeros and for an actual metric (defined as positive definite) there would be only ones. For a Lorentzian metric (one time-like direction), there would be one -1 and the rest of the diagonals would be 1.

George Keeling
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The canonical form of the metric has the same range of indices as any other form - four for relativity. Carroll is using ##-1,\ldots,-1## to mean "some number, possibly zero, possibly more, of -1s". In relativity the canonical form of the metric has three -1s, one +1, and no zeros (though sign conventions do vary!).

George Keeling
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I like #3 and #4 so I get $$\begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$ perhaps Carroll would have been clearer had he written $$g_{\mu\nu} = \rm{diag} (-1, 0,1,...-1, 0,1, ... -1, 0,1, ... ,-1, 0,1)$$ Thanks!

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That would have been wrong and very unclear.

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Does this work?
As a matrix, a non-degenerate metric in canonical form is diagonal with ±1 in each component.

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Does this work?
As a matrix, a non-degenerate metric in canonical form is diagonal with ±1 in each component.
Technically, that would be a pseudo-metric unless you have +1 in all diagonals, but in physics we just call it metric anyway.

George Keeling
$$\begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$
$$g_{\mu\nu} = \rm{diag} (-1, 1, 1,1)$$