Second fundamental form under conformal change

Question

I am calculating something in this paper https://arxiv.org/pdf/1501.07527.pdf at page 7, formula (10,11,12). Suppose $X:M^2\rightarrow (N,g)$ is an immersion, $\nabla$ is the connection on $N$ induced by $g$, $g^M$ is the induced metric on $M^2$. Now the second fundamental form of $X$ is defined to be $$A(Y,Z)=(\nabla_Y Z)^{\perp}.$$ The components $A_{ij}=(\nabla_{X_i}X_j)^{\perp}$, now I want to see what happens under conformal change of metric, i.e., $\tilde{g}=e^{2\varphi}g$, it goes like $$\begin{eqnarray}\tilde{A}_{ij}&=&(\tilde{\nabla}_{X_i}X_j)^{\perp}\\&=&(\nabla_{X_i}Y_i+X_i(\varphi)Y_i+Y_i(\varphi)X_i-g(X_i,Y_i)\nabla\varphi)^{\perp}\\ &=&(\nabla_{X_i}Y_i-g^{M}_{ij}\nabla\varphi)^{\perp}\\ &=&A_{ij}-g_{ij}^M(\nabla \varphi)^{\perp}\end{eqnarray}$$ Which is kind different from formula (12) which has $e^{\varphi}$ as coefficient in above paper, did I do something wrong?

Answer 1

tl;dr $A$ and $h_{i j}$ are different things and have different conformal weight.

Let us write $\overline{\nabla}$ for the intrinsic connection on $M$ (i.e., the Levi-Civita connection of $g^M$), and keep your notation $A$ for the second fundamental form:

$$ \nabla_Y Z =\overline{\nabla}_Y Z + A(Y,Z) $$

This is an orthogonal decomposition. In particular, $A$ is a vector-valued 2-form on $TM$. Actually, $A(Y,Z)$ takes values in the normal bundle to $M$.

In components, $A(Y,Z) = N \cdot h_{i j} Y^i Z^j$, where $N$ is a unit normal field along $M$. Unit normal vectors scale with the conformal weight $-1$, that is $\widetilde{N} = e^{-\varphi}N$. This follows from $g(N, N) = 1$ and the tautological fact that $g$ "scales" with conformal weight 2: $\widetilde{g} = e^{2 \varphi} g$, whereas the $1$ is weightless (a constant!).

Thus, $h_{i j}$ should have the conformal weight $1$, that is scale with the factor $e^{\varphi}$.

In the article the OP refers to, the authors swallow (p.6, between equations (8) and (9)) the transition from the vector-valued $h$ to the component scalars $h_{i j}$, and then later switch to an "associated endomorphism" like $[h]^{i}_{j}$, making the things even more confusing (cf. eq.(13)).