What is the canonical form of the metric?

[George Keeling]

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?

 

[martinbn]

He means exactly the same thing as the wiki article.

 

[Orodruin]

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.

 

[Ibix]

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]

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!

 

[Orodruin]

That would have been wrong and very unclear.

 

[George Keeling]

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

 

[Orodruin]

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.

 

[martinbn]

This
$$
\begin{pmatrix}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} $$
and this
$$ g_{\mu\nu} = \rm{diag} (-1, 1, 1,1) $$
are the same.