Difference between the metric tensor in general relativity and the metric tensor in mathematics?

Question

Is the metric tensor in general relativity the same as the metric tensor in maths, or is there a difference?

Answer 1

They are exactly the same thing.

The only difference is that one usually studies Riemannian metrics in mathematics, whereas the metric of relativity has a signature $(-+++)$, i.e. is not positive definite. Say the metric induces an inner product $\langle.,.\rangle$. For a positive definite metric, we have, for all vectors $V$ $$\langle V,V\rangle\ge0$$ with equality holding iff $V=\vec 0$. For a more general pseudo-Riemannian metric, we have $$\langle V,V\rangle>0$$ if $V$ is spacelike $$\langle V,V\rangle<0$$ if $V$ is timelike and $$\langle V,V\rangle=0$$ if $V$ is null. Note that a null vector must not be $\vec 0$, but $\vec 0$ is null. We often call the pseudo-Riemannian metric of GR a Lorentz metric.

The negative in the signature of the Lorentz metric changes some formulas from the Riemannian case. I know some examples off the top of my head.

In standard geometry, the invariant volume element is $\sqrt{g}dx^1\wedge\cdots\wedge dx^n$. In GR the determinant is negative, so we write $\sqrt{-g}dx^1\wedge\cdots\wedge dx^n$ to make the root real.

Let $\omega$ be a $p$-form on an $n$-dimensional manifold $M$. The application of the Hodge dual twice is $$*(*\omega)=(-)^{p(n-p)}\omega$$ In GR, this gets changed to $$*(*\omega)=(-)^{p(n-p)+1}\omega$$ with the extra minus arising from the signature of the metric.

Let $\Omega^p(M)$ be the space of $p$-forms on $M$. The map $$\delta:\Omega^p(M)\longrightarrow\Omega^{p-1}(M)$$ is called the codifferential and is defined by $$\delta=(-)^{n(p+1)}*d*$$ In GR this gets modified to $$\delta=(-)^{n(p+1)+1}*d*$$

There are many such sign changes due to the signature of the metric, but nothing really groundbreaking mathematically besides the existence of nonzero null vectors. (Physically, the metric signature plays a HUGE role in the study of causality. See Wald (1984) or Hawking & Ellis (1973) for a great coverage of causality in GR and Cosmology. Beware, the math is very formidable.)

Answer 2

This question was already answered very sufficiently, but I feel the need to clarify some other things, which may be trivial to you but maybe not.

Topologists like the refer to a function $$d:X\times X\rightarrow\mathbb{R}$$ where $X$ is a topological space, and the function is strictly positive (except when you plug in the same point twice, in which case it is 0) and satisfies triangle inequality, as a metric. It is basically a distance function.

This is obviously not the same thing as a metric tensor (to which geometers and relativists like to refer as metric), although you CAN derive a distance function from a metric tensor.

The topological metric will never be called metric tensor however.

Another fine point is, that technically an inner product on a real vector space could be called a metric tensor, since it lives in $V^{*}\otimes V^{*}$, so it is a tensor, but algebraists almost never refer to it as that. The primary difference between that and the geometers'/relativists' metric tensor is that the metric tensor is a tensor field that gives and inner product on all tangent spaces of a differential manifold, while an inner product is basically an algebraic tensor.

Other than that, as others have mentioned, geometers generally use positive definite metric tensors, while physicists use indefinite metric tensors, but this distinction is not a very strict one, since mathematicians also deal with pseudo-Riemann manifolds with indefinite metrics and physicists technically use euclidean metrics all the time (which are positive-definite), just most of the time that is not being spelled out, and even in relativity one often encounters spacelike hypersurfaces whose induced metrics are positive definite.