I have a question the variation of the Polyakov action, related to this Phys.SE post.
For Polyakov action $$ S_p[X,\gamma]=-\frac{1}{4 \pi \alpha'} \int_{-\infty}^{\infty} d \tau \int_0^l d \sigma (-\gamma)^{1/2} \gamma^{ab} \partial_a X^{\mu} \partial_b X_{\mu}.$$
Consider the variation (p. 14 in Polchinski's string theory book)
$$ \delta S_p~=~ \frac{1}{2\pi \alpha'} \int_{-\infty}^{\infty} d \tau \int_0^l d \sigma (-\gamma)^{1/2} \delta X_{\mu} \nabla^2X^{\mu}$$ $$ - \frac{1}{2\pi \alpha'} \int_{-\infty}^{\infty} d \tau (-\gamma)^{1/2} \delta X_{\mu} \partial^{\sigma} X^{\mu} |^{\sigma=l}_{\sigma=0} \tag{1.2.27} $$
The open-string and closed-string boundary conditions
$$\partial^{\sigma} X^{\mu}(\tau,0)=\partial^{\sigma} X^{\mu}(\tau,l)=0 \tag{1.2.28} $$ $$ X^{\mu}(\tau,l) = X^{\mu}(\tau,0), \partial^{\sigma} X^{\mu}(\tau,0)=\partial^{\sigma} X^{\mu}(\tau,l), \gamma_{ab}(\tau,l)= \gamma_{ab}(\tau,0) \tag{1.2.30} $$
will make the "surface term" (the second term in the RHS of Eq. (1.2.27)) vanishes. It is said (p. 14-15 of Polchinski)
The open-string boundary condition (1.2.28) and closed string boundary condition (1.2.30) are the only possibilities consistent with $D-$dimensional Poincare invariance and the equation of motion.
The boundary condition (1.2.28)/(1.2.30) will make the surface term vanish. The resulting equation of motion is tensor equation.
$$(-\gamma)^{1/2} \nabla^2 X^{\mu}=0$$
My question is, why non-vanishing surface term will break the Poincare invariance? E.g. a Lorentz transformation $X_{\mu} \rightarrow \Lambda_{\mu}^{\nu} X_{\nu}$ will not affect $\sigma,\tau$ space
