So I was reading through Becker, Becker, Schwarz and there is a line in the second chapter that states that $h^{\alpha \beta} = (h_{\alpha \beta})^{-1}$ where $h_{\alpha \beta}$ is defined as: $$h_{\alpha \beta} = \frac{\partial X^{\mu}}{\partial \sigma^{\alpha}} \frac{\partial X^{\nu}}{\partial \sigma^{\beta}} g_{\mu \nu}$$ where $X^{\mu}$ is our coordinates on our spacetime manifold, $\sigma^{\alpha}$ is our coordinates on our worldsheet, and $g_{\mu \nu}$ is our spacetime metric. This seems very natural given that $h_{\alpha \beta}$ is precisely the induced metric on the worldsheet and for metrics on our spacetime $g_{\mu \nu}^{-1} = g^{\mu \nu}$. However, I am having a hard time proving this. Namely, $$h_{\alpha \beta}h^{\alpha \gamma} = \frac{\partial X^{\mu}}{\partial \sigma^{\alpha}} \frac{\partial X^{\nu}}{\partial \sigma^{\beta}} g_{\mu \nu}\frac{\partial X^{\mu'}}{\partial \sigma_{\alpha}} \frac{\partial X^{\nu'}}{\partial \sigma_{\gamma}} g_{\mu' \nu'}$$ which doesn't look like it will yield $\delta_{\beta}^{\gamma}$. I have tried fiddling with the algebra with no avail...
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$\begingroup$ I think it's only the inverse once you impose ADM coordinates? I'm a bit busy to answer specifically, but this question is answered in most numerical relativity textbooks, which tend to be a bit gentler with this sort of thing than is otherwise typical. See Baumgarte and Shapiro for example. $\endgroup$– AGMLCommented Oct 2, 2019 at 17:13
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$\begingroup$ What's the definition of the coordinates $\sigma_{\alpha}$ with sub-indices (rather than super-indices)?? $\endgroup$– Qmechanic ♦Commented Oct 2, 2019 at 17:39
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7$\begingroup$ How do you define $\sigma_a$? I think $h^{ab}$ is simply defined to be the inverse of $h_{ab}$, no? $\endgroup$– user1379857Commented Oct 2, 2019 at 17:39
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$\begingroup$ This a definition. The question that perhaps you should ask instead is what compatibility condition allows this as a free-choice so that everything remains consistent between the two metrics. $\endgroup$– BrickCommented Oct 2, 2019 at 17:41
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1$\begingroup$ You are essentially projecting down to a lower-dimensional subspace. The compatibility should be with the elements of the original metric that "have legs" pointing out of that subspace, e.g. the lapse and the shift. Write $g_{ab} g^{ac}$ in terms of $h_{ab}$ and $h^{bc}$ plus lapse and shift. $\endgroup$– BrickCommented Oct 2, 2019 at 18:04
3 Answers
As you have stated, $h_{\alpha\beta}$ is the induced metric on the worldsheet, which we obtain by acting with what are analogous to projection operators on the metric.
This is a metric in its own right, and by definition the inverse in the matrix sense is $h^{\alpha\beta}$, such that, in $d$ dimensions, $h^{\alpha\beta}h_{\alpha\beta} = \mathrm{Tr} \, \mathbb I = d$.
It should be noted that $h$ is also often used for the fundamental form in most differential geometry literature which whilst related to the induced metric, carries spacetime indices still. The definition most useful to physicists, but not the most general, is that it is given by,
$$h_{\mu\nu} = g_{\mu\nu} \pm n_\mu n_\nu$$
where $n_\mu$ are the unit normals and the sign depends on if they are spacelike or timelike. This one is not as often used in string theory, but it is in general relativity, and I am pointing it out here so you don't get confused between the two.
You have to be a bit careful if you're working in the 4-geometry. I take the (-,+,+,+) metric, so the unit time/spacelike normal (the null case requires more care than the scope of this answer) is $n_{a}$ and $n_{a}n^{a} = \mp 1$
Then,
$$h_{ab} = g_{ab} \pm n_{a}n_{b}$$
obviously annhilates $n_{a}$ and just lowers any vector for which $n_{a}v^{a} = 0$
to find the inverse, take the ansatz:
$$h^{ab} = g^{ab} + C n^{a}n_{b}$$
we obviously want $h^{ab}n_{a} = 0$, so $C = \pm 1$
and then we just have to check:
$$h^{ab}h_{bc}=\left(g^{ab} \pm n^{a}n^{b}\right)\left(g_{bc} \pm n_{b}n_{c}\right)$$
Which, gives:
$$\delta^{a}{}_{c} \pm 2n^{a}n_{c} -\mp n^{a}n_{c} = \delta^{a}{}_{c} \pm n^{a}n_{c}$$
which is very obviously a projector operator which annhilates $n^{a}$. So that's the form for the 3-metric in the embedding in the 4-space
In differential geometry, $h$ is the pullback of the spacetime metric $g$ along the immersion of the worldsheet. Thus it is automatically a metric and writing this in local coordinates will give the formula you outline.
Like any metric, the composition of the dual metric and the metric itself gives the identity.
I'll write out the algebra explicitly later (once I find my notes).
