I have a super stupid question about deriving Eq. (1.2.17) in Polchinski's string theory, vol 1.
The book seems to derive from
$$\tag{1.2.16} h_{ab}=\frac{1}{2} \gamma_{ab} \gamma^{cd} h_{cd} $$
to
$$\tag{1.2.17} h_{ab} (-h)^{-1/2} = \gamma_{ab} (-\gamma)^{-1/2}.$$
I cannot figure out how to manipulate the determinants $\gamma$ and $h$. Excuse me, would you provide at least any hint for the derivation?
Calculate the determinant of (1.2.16). Note that these are determinants of $2\times 2$ matrices, as Peter Kravchuk told you, so $\det(CM)=C^2\det(M)$ for a constant $C$. Just to be sure, by the word "constant", I really mean $C$ is a scalar, not a matrix. This $C$ may still depend on $\tau,\sigma$ (or other variables if there were any).
There's a coefficient $C^2$ because two columns (or two rows) are multiplied by $C$ each and the determinant is multilinear in the rows (or columns). So the determinant of (1.2.16) is $$\det_{a,b} (h_{ab}) = \det_{ab} \left( \frac{1}{2}\gamma_{ab} \gamma^{cd}h_{cd} \right)$$ Because $\gamma^{cd}h_{cd}/2$ plays the role of the constant $C$ from my previous general identity, the equation above may be simplified to $$\det_{a,b} (h_{ab}) = \det_{ab} (\gamma_{ab})\times \left( \frac{1}{2} \gamma^{cd}h_{cd} \right)^2$$ Using the symbols such as $h=\det_{ab}(h_{ab})$ and similarly for $\gamma$, the square root of the minus equation above (minus because the determinants are negative in a Minkowskian signature) is $$\sqrt{-h} = \sqrt{-\gamma}\times \frac 12 \gamma^{cd}h_{cd} $$ The second power disappeared again. Divide (1.2.16) by the displayed equation I just wrote down. On the right hand side, $(1/2)\gamma^{cd}h_{cd}$ will cancel again and you're left with (1.2.17).
In none of the arguments above, it's important that $C$ may depend on $\tau,\sigma$. Also, it doesn't matter at all that/whether $h_{ab}$ may be calculated as the induced metric from $\partial_\alpha X^\mu$ etc.
Why others can derive the second equation immediately
Those experienced among us see the result immediately because the left hand side of (1.2.17) is the tensor proportional to $h_{ab}$ with a normalization constant chosen so that the determinant of this left hand side equals minus one (the minus can't be eliminated because it's given by the Minkowski signature). In other words, the normalization factor is chosen exactly to "forget" the normalization. Similarly, the right hand side of (1.2.17) is the tensor proportional to $\gamma_{ab}$ with the right normalization factor to makes the determinant of this product equal to minus one. Because both $h_{ab}$ and $\gamma_{ab}$ were converted – by adding $(-h)^{-1/2}$ factors etc. – to matrices whose determinant is equal to minus one, both sides of (1.2.17) actually encode the information about the matrices $h_{ab }$ and $\gamma_{ab}$ up to their normalization. In other words, (1.2.17) is equivalent to saying that $h_{ab}$ and $\gamma_{ab}$ are proportional to one another as matrices. The equation (1.2.16) also says that they're proportional to each other as matrices – with a coefficient that is explicitly given in (1.2.16) but not in (1.2.17) – so (1.2.17) follows from (1.2.16) (but not the other way around). This paragraph may sound complicated but it's really obvious for those who have a sufficient experience with tensors, matrices, and determinants.
Peter. Sorry I didn't get your point, $h_{ab} := \partial_a X^{\mu} \partial_b X_{\mu}$, $X^{\mu}(\tau,\sigma)$ is a function of the world-sheet. Therefore the determinant is a function of coordinate ${ \tau,\sigma }$, rather than constant $C$.
– user26143 Jul 09 '13 at 12:46