Vertex operator - state mapping in Polchinski's book In Polchinski's textbook String Theory, section 2.8, the author argues that the unit operator $1$ corresponds to the vacuum state, and $\partial X^\mu$ is holomorphic inside couture $Q$ in figure 2.6(b), so operators $\alpha_m^\mu$ with $m\ge0$ vanishes.
I am a bit confused about why $\partial X^\mu$ has no pole inside the contour. Before this section $\partial X^\mu$ always has the singularity part ($1/z^m$). Therefore would it be possible for you to give a more mathematical argument what condition requires $\partial X^\mu$ having no poles in this case?
 A: The main point is that the operator-state correspondence maps all the annihilation operators to zero, so that an operator-valued Laurent series in $z$ and $\bar{z}$ maps to a ket-state-valued power series in $z$ and $\bar{z}$.
A: Start with the action (2.1.1), where the fields $\partial X^{\mu}$ are defined on a given world sheet. Then eq. (2.1.12) tells you that $\partial X^{\mu}$ is holomorphic on this surface. 
The way we can develop a holomorphic function depends on the topology on the surface on which it is defined. If it is the complex plane (the usual case), then you have an expansion $\sum\limits_{m \geq 0} a_m z^m$. If you are on the Riemann sphere, you add a condition at infinity, and the only holomorphic functions (valued in the complex plane) are the constants. Now if you go on a cylinder, as is the case for the propagation of a free string, then you relax a condition at the origin, and the expansion is $\sum\limits_{m \in \mathbb{Z}} a_m z^m$. Even if this has "poles", it is indeed a holomorphic function on the cylinder, or the plane without the origin. 
Now in the state-operator correspondence, if you insert the identity operator at the origin, it reduces to going from the cylinder to the plane. There is no more hole at the origin, since now in picture 2.6.b the contour can cross the $\mathcal{A}$ insertion. So the coefficients with negative $m$ in $\sum\limits_{m \in \mathbb{Z}} a_m z^m$ must annihilate the corresponding state. This identifies this state with the ground state. 
A: Before we put a local operator at the origin of the $z$ plane, the origin is actually a singularity of the exponential map that sends the cylinder to the complex plane. So, the local operators are singular at the origin before the state-operator correspondence.
The state-operator correspondence is the observation that if we substitute the singularity at the origin with a local operator there, then we can map states at $t=\pm\infty$ on the cylinder to local operators at the origin of the plane.
The singular behavior at the origin is determined by the local operator that you put at the origin. The $1$ local operator is characterized by not having a singularity at all. Other states/operators like $:\partial X^{\mu}\bar{\partial}X^{\nu}e^{ikX}:$ reproduce a singularity at the origin.
A: I will give yet a different, simpler,  answer. All of these discussions are operator equations. Now as an operator equation $\bar{\partial}\partial X^\mu = \text{contact terms}$, see the discussion on p 35. But contact terms would arise from any operators inside the contour. The only such operator is the unit operator, and so there are no contact terms and $\bar{\partial}\partial X^\mu =0$ and so $\partial X^\mu$ is holomorphic.
