Why can we put these conditions on coordinates of worldsheet?

Question

https://www.asc.ohio-state.edu/mathur.16/classicalstring.pdf

At first, I write some notations I need here.

$I=[0,1]$, $M$ means $(1,3)$ Minkowski space, smooth map $X:I\times I\to M$ is timelike worldsheet and we denote $\tau$ as the first parameter of $X$ and $\sigma$ as the second parameter, $g^{ind}$ is induced metric by $X$.

Page 4 of this pdf, we can put the conditions by reparametrizing $X$

$g^{ind}_{\tau\sigma}=0 $

$g^{ind}_{\tau\tau}+g^{ind}_{\sigma\sigma}=0$

I don't understand why we can put these conditions. This pdf explain it by just one sentence "We can set two combinations to chosen values, by using the two freedoms of coordinates". I think "the two freedoms of coordinates" means that we can parametrize $X$ by arbitrary two functions of $\tau$ and $\sigma$. I name these functions as $\tilde{\tau}(\tau,\sigma)$ and $\tilde{\sigma}(\tau,\sigma)$ for each. By this parametrization, a component of metric is

$g^{ind}_{\tilde{a}\tilde{b}}=g^{ind}_{cd}\dfrac{\partial c}{\partial \tilde{a}}\dfrac{\partial d}{\partial \tilde{b}}$ ...(i) (Here, each a,b,c,d is $\tau$ or $\sigma$ and indices are contracted)

How do we know if it's possible to get new parametrization which satisfies the two conditions

$g^{ind}_{\tilde{\tau}\tilde{\sigma}}=0 $

$g^{ind}_{\tilde{\tau}\tilde{\tau}}+g^{ind}_{\tilde{\sigma}\tilde{\sigma}}=0$

?

In other words, are there solutions of this type of 2 differential equations always?

Answer 1

Such a choice of coordinates is equivalent to writing the induced metric as $$g^{ind}_{ab}(x)= h(x) \eta_{ab}$$ where $h$ is some scalar function on the worldsheet and $\eta$ is the 2d Minkowski metric. A metric for which this choice of coordinates exists is called conformally flat. All 2d manifolds are conformally flat, i.e. they can always be put in this form at least in a coordinate patch. I'll show how the coordinates can be constructed below:

Take a coordinate patch $U$ on the worldsheet and specify an arbitrary coordinate $\tau$ on the boundary $\partial U$. You can extend $\tau$ to the interior of $U$ by solving the Laplace equation $d\star d\tau=0$, or in coordinates $$\nabla_\mu \left(g^{\mu\nu}\partial_\nu \tau\right)=0.$$ Now let's consider the dual one form fields $$v_\mu \equiv \partial_\mu \tau,\qquad \tilde{v}_\mu\equiv \epsilon_{\mu\nu}v^\nu.$$

Using the properties of the Levi-Civita tensor we have $$g_{ind}^{\mu\nu}v_\mu\tilde{v}_\nu =0,\qquad g^{\mu\nu}_{ind}v_\mu v_\nu = -g^{\mu\nu}_{ind}\tilde{v}_\mu\tilde{v}_\nu.$$ So if we can say $\tilde{v}_\nu =\partial_\nu \sigma$ for some function $\sigma$ then these expressions above are exactly the coordinate transformation to put the metric in a manifestly conformally flat form.

And we can find such a function $\sigma$ since the exterior derivative of $\tilde{v}$ vanishes (this is the curved space version of the idea that a vector field with vanishing curl can be expressed as a gradient.) $$\epsilon^{\mu\nu}\nabla_{\mu}\tilde{v}_\nu=\epsilon^{\mu\nu}\epsilon_{\nu\rho}\nabla_{\mu}\left(g^{\rho\sigma}\partial_\sigma \tau\right)=\nabla_{\rho}\left(g^{\rho\sigma}\partial_\sigma \tau\right)=0 $$ which implies $\tilde{v}_\nu=\partial_\nu \sigma$ for some coordinate $\sigma$ at least on small enough $U$.

Answer 2

Only a small remark complementing the existing answer:

For Lorentzian surfaces (2D manifolds with $(-+)$ metric signature) there is a simpler proof of conformal flatness that does not require any serious result from PDE theory, only the Frobenius integrability theorem. The metric can always be written in terms of an ortonormal coframe as $$ ds^2=\theta_1^2-\theta_0^2, $$which is of course true for all metrics. But this expression can be factorized as $$ ds^2=(\theta_1+\theta_0)(\theta_1-\theta_0)=\theta_+\theta_- . $$

Since $M$ is two dimensional, every $1$-form satisfies $d\theta\wedge\theta=0$, which by the Frobenius theorem implies that there are functions $\alpha$ and $\phi$ such that $\theta=\alpha d\phi$.

Hence, there are functions $\alpha_+,\alpha_-,\phi_+$ and $\phi_-$ such that $\theta_\pm=\alpha_\pm d\phi_\pm$, thus$$ ds^2=\alpha_+\alpha_-d\phi_+ d\phi_-.$$ We may assume without loss of generality that $\alpha_+\alpha_-$ is positive and parametrize them as $\alpha_+\alpha_-=e^{2\psi}$. Define then the functions $t,x$ via $$ \phi_+=x+t,\quad\phi_-=x-t, $$which gives $$ ds^2=e^{2\psi}(dx+dt)(dx-dt)=e^{2\psi}(dx^2-dt^2). $$

For positive definite signature this proof does not work because the metric cannot be factorized as above and one does need some results from PDE theory concerning elliptic differential operators.

A proof for the positive definite case can be found in DeTurck, Kazdan: Some regularity theorems in Riemannian geometry and this proof is along the lines indicated by octonion .