4
$\begingroup$

Disclaimer. I am a mathematician (algebraic geometer) who knows nothing about physics. Even worse, I might have major misconceptions about the objects I'll ask about. The level of the question is pop-science, but perhaps just a bit too specific for something like Quora. I apologize in advance for my ignorance.

According to my understanding, a worldsheet is the 2D surface traced out by a vibrating string. We can think of this as (the image of) a map from a Riemann surface into spacetime. However, my understanding is that this map ought to be holomorphic. [Here I am already a bit confused, though this is not the main point of the question: do we only consider maps into a (6D) Calabi-Yau fiber of spacetime (are these called instantons?), or does this somehow make sense for maps whose image isn't contained in one of these fibers?]

My question: why does it have to be holomorphic?

How to come to terms (technically or philosophically) with this very strong hypothesis/property?

$\endgroup$
2
  • 1
    $\begingroup$ I don't think the map needs to be holomorphic in general. Do you have a reference that says this? $\endgroup$
    – d_b
    Commented Sep 24, 2019 at 7:27
  • $\begingroup$ Maybe what you mean is that the worldsheet equations of motion separate solutions in holomorphic and antiholomorphic parts? (See e.g. arxiv.org/abs/hep-th/9709062) $\endgroup$
    – Toffomat
    Commented Sep 24, 2019 at 10:58

1 Answer 1

2
$\begingroup$

I suggest learning the basics of conformal field theory before diving into string theory. To illustrate where holomorphicity comes in, let's consider the free bosonic string (the simplest example of a CFT). One starts with the worldsheet embedding $\phi(z,\overline{z}): \mathbb{C}P^1 \to \mathbb{C}$ and derives the equations of motion $\partial_{\overline{z}} \partial_z \phi(z,\overline{z}) = 0$. This implies the existence of the fields \begin{equation} \partial \phi(z) = \sum_{n \in \mathbb{Z}} a_n z^{-n-1} \ \ \ \ \ \ \ \ \ \ \overline{\partial} \phi(\overline{z}) = \sum_{n \in \mathbb{Z}} \overline{a}_n \overline{z}^{-n-1}, \end{equation} which are holomorphic and antiholomorphic respectively. So one answer to your question is that the fields of interest are holomorphic/antiholomorphic as a result of the equations of motion (which in the case of the free boson are derived from the action of the theory).

There is much more to be said here. To obtain a quantum theory, the coefficients $a_n$ are promoted to elements of a Lie algebra. This is known as canonical quantisation. In the case of the free boson, the Lie algebra is generated by the elements $\{a_n\}_{n \in \mathbb{Z}}$ and $\{k\}$, with Lie bracket defined by \begin{equation} [a_n, a_m] = \delta_{n, -m}k \ \ \ \ \ [a_n, k] = 0. \end{equation} There is another Lie algebra of interest known as the Virasoro algebra. The Virasoro algebra appears in 2d CFT as a result of conformal invariance of the theory. The Virasoro algebra has generators $L_n$ and one obtains another holomorphic field \begin{equation} T(z) = \sum_{n \in \mathbb{Z}} L_n z^{-n-2} \end{equation} known as the energy-momentum tensor. There is also an antiholomorphic analogue of the energy-momentum tensor. For this reason, we say the free boson is composed of holomorphic and antiholomorphic "sectors".

If you are interested in learning more about CFT, check out Conformal Field Theory by Sénéchal, Di Francesco, and Mathieu. For a more mathematically rigorous account, see Vertex Algebras and Algebraic Curves by Frenkel and Ben-Zvi (your background in algebraic geometry will be useful here).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.