# How to understand worldsheet fermion as a section?

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I am reading Witten's paper on topological string, and I found some mathematical notation is hard to understand for me. Consider the nonlinear sigma model in 2 dimensions governed by maps $\Phi : \Sigma \rightarrow X$ with $\Sigma$ being a Riemann surface and $X$ a Riemann manifold of metric $g$. $z, \bar{z}$ are local coordinate on $\Sigma$ and $\phi^I$ is coordinate on $X$. $K$ and $\bar{K}$ are canonical and anticanonical line bundles of $\Sigma$ (the bundle of one forms of types (1,0) and (0,1) repectively), and let $K^{1/2}$ and $\bar{K}^{1/2}$.The fermi fields of the model are $\psi_{+}^I$, a section of $K^{1/2}\otimes\Phi^*(TX)$.

I can not understand the sections of $K^{1/2}$, $\Phi^*(TX)$ and $K^{1/2}\otimes\Phi^*(TX)$.

From my point of view, the element of $K$ should be of the following form $\alpha_z dz\in K$, and what is the element of $K^{1/2}$? the pull back of tangent space should be of form $\Phi^*(\beta^i \frac{\partial}{\partial \phi^i})=\beta^i \frac{1}{\frac{\partial \phi^i}{\partial z} }\frac{\partial}{\partial z}$. But in some notes the author seems give that the (0,1) form $\psi_-^i$ with values in $\Phi^*(T^{1,0} X)$ can be written as $\psi_{\bar{z}}^i$ satisfying $\psi \supset \psi_{\bar{z}}^id\bar{z}\otimes \frac{\partial}{\partial \phi^i}$. This contradicts with my naive point of view. where did I make mistakes? How to understand the sections of $K^{1/2}$, $\Phi^*(TX)$ and $K^{1/2}\otimes\Phi^*(TX)$?

This post imported from StackExchange Physics at 2014-04-18 16:22 (UCT), posted by SE-user Craig Thone
In addition to the other answers, I should point out that one needs to be careful when reading that paper to pay attention to whether Witten is discussing a QFT or its associated "twisted" QFT. The latter will not involve square roots of the canonical bundle. You can confuse yourself by trying to match formulae from the twisted and untwisted theories.

This post imported from StackExchange Physics at 2014-04-18 16:22 (UCT), posted by SE-user user1504

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In general, the canonical bundle $K$ is the bundle of n-forms on an n-dimensional manifold. Since your Riemann surface $\Sigma$ is one (complex) dimensional, it's just the (line) bundle of holomorphic one-forms. The "square root" $K^{\frac{1}{2}}$ is the bundle of things which transform in a way which is sort of the square root of the transformation of the holomorphic one forms. So if, under a worldsheet coordinate transformation (where $z$ is a local coordinate on $\Sigma$) $$z\rightarrow e^{i\alpha}z$$a one-form transforms as$$\omega\rightarrow e^{i\alpha}\omega$$ then, for the square root, we want something transforming as $$\psi\rightarrow e^{i\frac{\alpha}{2}}\psi$$ This is just a right handed worldsheet spinor. If it's a RH one, it's denoted $\psi_{+}$ and if, instead, it transformed with a $-\frac{\alpha}{2}$, it's LH and denoted $\psi_{-}$

Now you also want your entity to take values in the bundle $\phi^{*}(TX)$.

$\phi:\Sigma\rightarrow X$ is an embedding. Thinking of $\phi^{i}$; $i=1..N$ (where $N$ is the dimensionality of $X$) as coordinates on $X$, then your desired object has a target space index. So for example the right handed version of the components would be $\psi^{i}_{+}$, and the section is $$\psi^{i}_{+}(z)\frac{\partial}{\partial \phi^i}$$

This post imported from StackExchange Physics at 2014-04-18 16:22 (UCT), posted by SE-user twistor59
answered Mar 31, 2013 by (2,500 points)

This post imported from StackExchange Physics at 2014-04-18 16:22 (UCT), posted by SE-user Craig Thone
$\psi_+^i(z)\frac{\partial}{\partial\phi}$ seems take values in $TX$ rather than $\Phi^*(TX)$, why not some form of $\psi_+^i \frac{1}{\frac{\partial \phi^i}{\partial z} }\frac{\partial}{\partial z}$? What is the difference of the sections of $K^{1/2}$, $\Phi^*(TX)$ and $K^{1/2}\otimes\Phi^*(TX)$?

This post imported from StackExchange Physics at 2014-04-18 16:22 (UCT), posted by SE-user Craig Thone
If I have a vector bundle $V$ over $X$, then the fibre of the pullback bundle $\phi^{*}(V)$ over $p \in \Sigma$ is just the fibre of $V$ over $\Phi(p)$, so since $\Phi$ here is an embedding, I can just work with the image of $\Sigma$ and restrict the bundle in question (in this case the tangent bundle) to this image.

This post imported from StackExchange Physics at 2014-04-18 16:22 (UCT), posted by SE-user twistor59
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Spinors are sections of the spinor bundle. The decomposition given by Witten of the spinor bundle is valid on Kähler manifolds which Riemann surfaces constitute special cases of. The spinor bundle has a structure group $SO(2N)$, where $N$ is the complex dimension of the Kähler manifold $M$. This group reduces to U(N) due to the existence of a Kähler structure. (This means that when one writes the curved space Dirac equation on $M$, it has the form of a gauged Dirac equation in flat space coupled to a U(N) gauge field).

The space of sections of the spinor bundle belongs a $2^N$ dimensional spinor representation of $SO(2N)$. This representation decomposes into a direct sum of all anti-symmetric representations of the $SU(N)$ factor of $U(N)$ under the reduction of the structure group. Thus from the dimension counting point of view, the $2^N$ dimensional spinorial representation space is isomorphic to the exterior algebra $\Lambda^{(0, *)}(M)$ of the holomorphic tangent bundle.

The appearance of the square root of the canonical bundle is due to the fact that the contribution of the $U(1)$ factor of $U(N)$ group to the Dirac equation is an Abelian connection whose curvature is just $\frac{N}{2}$ times the Kähler structure. This part provides the spin $\frac{1}{2}$ character of the fermion fields.

This post imported from StackExchange Physics at 2014-04-18 16:22 (UCT), posted by SE-user David Bar Moshe
answered Mar 31, 2013 by (4,355 points)

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