# Readings

###### tags: `reading` `strings` `qft`

### Polchinski 1.1-1.2
We get introduced to the action and the symmetries. The *tetrad* formulation is useful to get rid of the square roots.

* Explanation of eq(1.2.17) : $\det \alpha A = \det \alpha \det A = \alpha^2 \det A$ when $\alpha$ is a constant and we are dealing with 2 dimensions. Same reason for the explanation below eq(1.2.21).

* Explanation of eq(1.2.23) : This is because under scaling (infinitesimal) $\gamma_{ab}\rightarrow e^{2\omega} \gamma_{ab} \simeq \gamma_{ab} + 2\omega \gamma_{ab}+\dots$ we find that $\delta\gamma_{ab}\propto \gamma_{ab}$. Hence the action of deformation along the direction of symmetry is: $\delta\gamma_{ab}\frac{\delta{S}}{\delta\gamma_{ab}} \propto \gamma_{ab}\frac{\delta{S}}{\delta\gamma_{ab}}$.

* Interesting way to define internal symmetry in page 14 : a symmetry which acts *locally*. Why cannot there be non-local internal symmetries?

* The symmetry property : tracelessness of stress tensor is also true off-shell. And on shell the stress tensor vanishes (for classical configurations): <span style="color:blue">This property is only true in String theory not in general CFTs, where classically too one may have a non-trivial stress tensor?</span> Seeking clarification on this point.

* $\int d\tau \int_0^\ell d\sigma \sqrt{-\gamma} \nabla_a \left( \delta X \nabla^a X\right)$ Adding onto the String action: $\int d^2\sigma \sqrt{-\gamma} A$ : $\delta \sqrt{-\gamma} \propto \sqrt{-\gamma} \gamma^{ab} \delta \gamma_{ab}$ : $\partial_a \partial_b$ can contract the $\gamma^{ab}$. $A$ is therefore made out of two worldsheet derivatives.

* Something similar to eq(1.2.32) is worked out in Problem 3 of [phy421-2019-endsem](http://home.iitk.ac.in/~didas/readings/2019-421-endsem.pdf)

### Polchinski 1.3 till eq(1.3.21)

Polyakov action has unecessary redundancies at the expense of no square root. Hence we need to gauge fix. *Quickest* way to do this is by going to the light cone coordinates.

* Consistent with the gauge choices, $\sigma$ is identified with the length on the string. This condition already implies that $\partial_\sigma f =0$. Next, with the explicit definition of $f$ and the condition 1.3.8c we get $\partial_\sigma \gamma_{\sigma \sigma}=0$. It is unclear why one needs Weyl invariance of $f$ as an intermediary argument.

* <span style="color:blue">We may need little more discussion on footnote 4.</span>

* Condition 1.3.14 : $\gamma_{\tau \sigma}= 0$ at the endpoints for all values of $\tau$. Next the equation of motion of $Y^-$ implements the constraint: $\partial_\sigma \gamma_{\sigma \tau}=0$ for all $\tau$. Hence we can conclude, that also away from the endpoint $\gamma_{\tau\sigma} = 0$, hence vanishes identically (*to remember this is something satisfied by constraints, in general for instance when one solves for constrained instantons, one may not always implement the constraint equation before quantization!*)

### Polchinski 1.3 till eq(1.3.31)

We mainly got us all upto speed. Next up, we plan to finish the rest of the first chapter and as well as attempt a few problems.

* Problem 1.5 : $$\sum_{n=1}^\infty (n-\theta) = \sum_{n=1}^\infty (n-\theta)\exp\left[ -\epsilon \gamma_{\sigma \sigma}^{-1/2} | (n-\theta)\pi/ \ell |\right]$$
$$=\sum_{n=1}^\infty (n-\theta)\exp\left[ -\epsilon \, \left(\frac{2\pi \alpha'p^+}{\ell}\right)^{-1/2} | (n-\theta)\pi/ \ell |\right]$$
Next assuming ($\theta\leq 1$) we use the formula:
$$\sum_{n=1}^\infty (n-b) \exp \left( -a (n-b)\right) = \frac{e^{ab}(b + e^a - b e^a)}{(e^a-1)^2 }$$
we expand for small $\epsilon$ to obtain:
$$
\sum_{n=1}^\infty (n - \theta ) \rightarrow c \frac{\ell}{\epsilon^2} + {\mathcal{O}}(\epsilon) - \frac{1}{12} - \frac{\theta(\theta-1)}{2}.$$
The cut-off independent finite part is the regularized answer thus:
$$
\sum_{n=1}^\infty (n - \theta) = - \frac{1}{12} - \frac{\theta(\theta-1)}{2} = \frac{1}{24}- \frac{(2\theta-1)^2}{8}.
$$
The last expression indeed matches with eq(2.9.19).