# Why does product of Moduli and Diff x Weyl Variation vanish?

According to equation 5.2.5 in Polchinski :-

$$\int d^2 \sigma~ \delta^{'}g_{ab} \times [-2(P_1 \delta \sigma)_{ab} +(2\delta w - \Delta \cdot \delta \sigma)g^{ab}]=0$$

The assumption here is that " Moduli correspond to variation $\delta^{'}g_{ab}$ of the metric that are orthogonal to all variations of DiffxWeyl type given by the quantity in []"

My question is that why are we imposing that orthogonality ? How can variation in the metric due to moduli be orthogonal to variation due to other Diff x Weyl ?

You may think to a inner product $(T, T')$:

$(T,T') = \int d^2\sigma \sqrt{g} \,T. T'$

where $T$ and $T'$ are tensors of equal rank, and $T.T'$ corresponds to a contraction on the tensor indices.

Now, your expression is simply, that, for all the Diff-Weyl variations $\delta g_{DW}$, and for all the moduli variations $\delta g_M$, you have :

$(\delta g_M, \delta g_{DW})= 0$

So the space of the moduli variations is orthogonal to the space of the Diff-Weyl variations (you may think at "vectorial" spaces).

• Thanks for the clarification but is there any reason to take the space of moduli variation to be orthogonal ? – sol0invictus Aug 28 '14 at 16:45
• @sol0invictus : It is, in some sense, the definition of the moduli, the moduli space (see $5.2.2$) is the "quotient" of space metrics by Diff Weyl redundancy. For instance, the quotient of $SO(4)$ by $SU(2)$ is $SU(2)$ because the two $SU(2)$ are "orthogonal" (independent). But the quotient of $S^3$ by $S^2$ is not $S^1$. There is a local fiber $S^1$, but it is not a global one, and there is not global $S^1$ and $S^2$ which are orthogonal. – Trimok Aug 29 '14 at 8:40