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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?

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2 Answers 2

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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 .

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  • $\begingroup$ Thank you for answering! $ds^2=(\theta_1+\theta_0)(\theta_1-\theta_0)$ how do you get this? In other words, $θ_1θ_0=θ_0θ_1$ is true? $\endgroup$ Commented Dec 22, 2022 at 9:31
  • $\begingroup$ @particle Depending on whether you take the metric as a bilinear form or a quadratic form, the product $\theta\omega$ of two $1$-forms is either defined to be the symmetric tensor product $\theta\omega=(1/2)[\theta\otimes\omega+\omega\otimes\theta]$ or the "pointwise product" $(\theta\omega)(X)=\theta(X)\omega(X)$ valid for any vector field $X$. Both are commutative products and both reproduce the desired result. $\endgroup$ Commented Dec 22, 2022 at 14:15
  • $\begingroup$ Thanks for reply! "$dθ∧θ=0$, which by the Frobenius theorem implies that there are functions α and ϕ such that $θ=αdϕ$." Sorry I dont understand which Frobenius theorem you used. Could you write the assumption and claim of the theorem? $\endgroup$ Commented Dec 23, 2022 at 7:06
  • $\begingroup$ And I also want to know if I can use the coordinates for a whole world sheet. $\endgroup$ Commented Dec 23, 2022 at 7:31
  • $\begingroup$ @particle This Frobenius theorem: en.wikipedia.org/wiki/… . Might update the answer later with more details. Also it is unlikely that the isothermal coordinates can cover the whole worldsheet unless the worldsheet has trivial topology. For closed strings this is certainly not the case. $\endgroup$ Commented Dec 23, 2022 at 15:39
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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 $U$.

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  • $\begingroup$ Thanks for answering!! >>"All 2d manifolds are conformally flat" Then, for each 2d surface which is given metric and homeomorphic to $I\times I$, we can choose coordinates in which the metric is expressed as h(x)η_{ab}, right? >> "timelike gradient dτ" Does that mean g^{ττ} is negative? >>"By construction ∗dτ is orthogonal to dτ..." Yes, I just checked. >>"This is just Laplace's equation associated to the metric g^{ind}..." Could you explain it in more detail or introduce any website in which I can confirm d( ∗dτ)=0? $\endgroup$ Commented Dec 21, 2022 at 16:30
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    $\begingroup$ You can't confirm $d(*d\tau)=0$ for an arbitrary function, the idea is that you have to choose a function $\tau$ that solves this equation. I'll put up an alternate derivation in an edit. $\endgroup$
    – octonion
    Commented Dec 21, 2022 at 20:23
  • $\begingroup$ >"I'll put up an alternate derivation in an edit" --- Thank you very much!! I also want to know about your sentence "All 2d manifolds are conformally flat, i.e. they can always be put in this form at least in a coordinate patch". Can we put parameters which gives expression of metric, $g^{ind}_{ab}(x)=h(x)η_{ab}$ to whole a 2d manifold which is diffeomorphic to$I\times I$? And I'd like to also know the proof of the fact "All 2d manifolds are conformally flat" Is there any website show the proof? $\endgroup$ Commented Dec 21, 2022 at 21:17
  • $\begingroup$ Actually my answer was intended to be a sketch of a proof of that by explicitly constructing the coordinates. I rewrote it (using components rather than differential forms too). $\endgroup$
    – octonion
    Commented Dec 22, 2022 at 5:19

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