A question about surface term of ghost fields (skip disclaimer)
Hi, I have a question in Polchinski's string theory vol I p 90, after introducing the ghost fields $b_{ab}$ and $c^a$, it is claimed

The equations of motion then provide a boundary condition on $b_{ab}$. They have a surface term
  $$ \int_{\partial M} ds n^a b_{ab} \delta c^b=0 \tag{3.3.28}$$

I simply don't know where does it comes from. There are some boundary condition
$n_a c^a=0 (3.3.26)$ but with different indices. How to see Eq. (3.3.28) holds?
 A: In order to get this term, you have to apply partial integration with respect to one of the coordinates to the action. This leaves you with one integral over two dimensions (with the derivative switched from $c$ to $b$) and an integral over one dimension. If you now vary the second term with respect to $c$ and you acquire Eq. 3.3.28.  
A: Partial answer : 
I think that the problem, in Polchinski book, is that the terms $g^{z \bar z} (=2)$, are systematically non explicitely written, that is, instead of : 
$\frac{1}{4 \pi}g^{z \bar z}$, Polchinkski write : $\frac{1}{2 \pi}$
So, for instance, the correct expression in $(3.3.24)$ is :
$$ S = \frac{1}{4 \pi} \int d^2z (g^{\bar z z}b_{zz} \partial_\bar z c^z + g^{z \bar z }b_{\bar z \bar z} \partial_z c^\bar z)$$ 
Now, because $b_{z \bar z}=0$, because $b$ is traceless,and $g^{zz}=g^{\bar z \bar z}=0$ ,this is really an invariant, that is : 
$$ S = \frac{1}{4 \pi} \int d^2z (g^{ i j}b_{jk} \partial_ i c^k)$$, 
where ($i,j = z, \bar z$)
Now, an invariant is an invariant, so you could write it with ($i,j = \sigma^1, \sigma^2$), if, you want, so, with Polchinski notations : 
$$ S = \frac{1}{2 \pi} \int d^2 \sigma (g^{ ca}b_{ab} \partial_c c^b)$$, 
Then the boundary term appears as : 
$$\delta B \sim \int ds~ g^{ ca} n_c ~b_{ab}  ~\delta c^b = \int ds ~  n^a ~b_{ab}  ~\delta c^b $$
