Re: GR1916 question about g=1?

John C. Polasek wrote:
the metric is ds^2 = sum of squares. The metric is ds^2 =
gijdqidqj.

A nit, but an important one in this discussion: that is NOT the metric, that is the line element. The line element expresses the length of a differential element (ds) of a line in terms of the coordinate differentials of the element and the metric COMPONENTS projected onto that coordinate basis.

This is something that Koobee does not understand at all,
and Tucker is rather vague about. Daryl understands it
quite well.

Yes, some less-than-careful authors call that "the metric", which is really an abuse of terminology. Especially older authors (pre-1970 or so).

[See the "For advanced readers only" box below. It's
clear to me that the alternate interpretation given
there is not what Polasek has in mind.]


The matrix is the metric "tensor".

NO! That matrix is composed of the metric COMPONENTS projected onto some specific coordinate basis.

Note that we often use the shortcut "the metric" to abbreviate the phrase "the metric tensor function on the manifold".

A tensor is not at all a matrix. A rank-N tensor is a multi-linear map of N vectors on to the reals (occasionally other-valued tensors are used in very advanced discussions, such as spinor-valued 1-forms).

Because it is multi-linear, once one selects a basis for
the vectors and expresses the vectors as components wrt
the basis, then a rank-2 tensor can be expressed as a
matrix of its components projected onto that basis.
This is what is done in the line element above. The
actual tensor can be expressed as the tensor product of
this component matrix and the covectors of the basis.
As that matrix is a set of real numbers, it should be
obvious that the tensor itself is thus a sum of tensor
products of pairs of covectors -- that cannot possibly
"be" a matrix, as those covectors are elements of the
cotangent space of the manifold....

[For advanced readers only: there is an alternative
interpretation of the line element equation:
ds^2 = g_ij dx^i dx^j
in which the {g_ij} remain the components of the metric,
but the {dx^i} are interpreted as the set of 1-forms
corresponding to the coordinate basis, and the tensor
product symbol between them has been omitted. Now that
equation defines a rank-2 tensor (those 1-forms are
the "covectors" mentioned above) -- this then _IS_
the metric tensor, but "ds^2" seems like a rather
strange symbol for it.... Omitting the tensor product
symbol makes this interpretation unpalatable.]


The matrix is not the metric. It's not the metric. It's not the metric.

Right.


It's the metric matrix. The metric is ds^2 = gijdqidqj.

Wrong on both. See above for the correct usage of these terms.

This is all very basic analysis on manifolds. There is
no "debate" on any of these terms or their meanings.
But some people around here are highly confused and
do not understand either the proper usage or the
underlying concepts.


Tom Roberts
.

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