I am trying to understand the Dirac operators associated to the 2 spinor bundles on $S^1.$ I have been getting very confused about why one bundle has nontrivial harmonic spinors and the other doesn't.(Harmonic spinors are solutions $s$ to the equation $Ds = 0$ where $D$ is the Dirac operator and $s$ is a section.)

Here is my argument (which must be wrong somewhere). We have 2 spin structures given by the connected 2-fold covering and the disconnected 2-fold covering. Since the tangent bundle $TS^1$ is trivial, we can choose the trivial connection on it given by $f \rightarrow df.$ When considered as a connection on the principal bundle of frames (also isomorphic to $S^1$), i.e. as a Lie algebra valued one form on $S^1,$ it must be the zero form.

(As a quick aside, the Lie algebra of $SO(1)$ is just the $0$ Lie algebra, so it seems like there is only one connection on the tangent bundle of $S^1$ since we could only ever have the $0$-form as the connection form on our frame bundle. But this is not true, we can add a 1-form to any connection and get another connection. How can this be?)

(EDIT: Answer provided by Eric: because we have implicitly reduced the structure group of the frame bundle to $SO(1),$ an $so(1)$ valued one form corresponds to a connection compatible with the given metric, and there is only one of these since the torsion of any connection on $S^1$ is zero.)

Ok, so now given either spin structure, the connection must lift to the $0$ connection. Furthermore, any complex line bundle over the circle is trivial, so both cases look exactly the same, and the Dirac operator appears to be $f \rightarrow i\frac{df}{dx}.$

However, I am told that in the case of the connected double cover we should have an additional condition on our $f,$ namely that it should satisfy $f(-x) = -f(x).$ Where have I gone wrong?

(2nd EDIT) I think I know where my confusion stems from. Given a spin structure $P$ on a manifold $M^n$ we can identify sections of the spinor bundle with smooth maps $f: P \rightarrow \mathbf{C}^n$ such that $f(pg) = g \cdot f(p)$ as follows. Take any discontinuous section $s: M \rightarrow P.$ This gives a section $t:M \rightarrow P \times_{Spin(n)} \mathbf{C}^n$ of the spinor bundle by the formula $t(m) = [s(m), f \circ s(m)],$ and by the compatibility condition on $f$ this is independent of the choice of section $s.$

It is via this identification that I understand how sections of the spinor bundle for the connected double cover must be functions $f: S^1 \rightarrow \mathbf{C}$ such that $f(-x) = f(x).$ And, as luck would have it, in this case the Spin structure on $S^1$ is itself, as a space, $S^1,$ so the description of the Dirac operator translates easily via this identification. But in the case of $S^2,$ for example, sections can be identified with maps from $S^3$ to $\mathbf{C}^2$ but the local description of the Dirac operator I see in books (like Lawson/Michelson's Spin Geometry) tells me how to differentiate maps $U \subset S^2 \rightarrow \mathbf{C}^2$ in a trivializing nbhd $U.$ How do I translate the local description of the Dirac operator so that it tells me how to differentiate the map $S^3 \rightarrow \mathbf{C}^2$?

  • $\begingroup$ I'm a little confused about what these spin structures are. Usually a Spin structure is a principal $Spin(n)$ bundle that covers the frame bundle such that restricting to each fiber gives the double cover $Spin(n) \to SO(n)$. Here $SO(1)$ has no double cover since it's simply connected. Are you saying that the covers of $S^1$ give spin structures? $\endgroup$ Commented Dec 1, 2012 at 5:31
  • 1
    $\begingroup$ With regards to your question about the connection, notice that only connections that preserve the metric have connection one-forms with values in $SO(1)$. And in the case of $S^1$, the Levi-Civita connection is the only connection that preserves the metric since the torsion of any connection is identically zero. $\endgroup$ Commented Dec 1, 2012 at 5:32
  • $\begingroup$ Eric: Yes, in this case $Spin(1) = \mathbf{Z}/2\mathbf{Z}$ and one of the double covers of the frame bundle is the unique connected double cover over $S^1$ by $S^1$ where the map is $z \rightarrow z^2.$ The other is given by the disconnected double cover, $S^1 \amalg S^1.$ $\endgroup$
    – mck
    Commented Dec 1, 2012 at 13:03
  • $\begingroup$ I do not really see why what you wrote in the second edit is true. For example, I see no reason why your sections should be smooth. $\endgroup$
    – Kofi
    Commented Dec 2, 2012 at 9:17
  • $\begingroup$ I'm getting this from Friedrich's "Dirac Operators in Riemannian Geometry." He claims that the correspondence holds and gives no proof, however I think that what I've said is right and the map will be smooth so long as the map $P \rightarrow \mathbf{C}^n$ is smooth. In any case, I think that this must just be an unhelpful way to think about sections of the spinor bundle, and I'm going to close the topic. Thank you for your help! $\endgroup$
    – mck
    Commented Dec 2, 2012 at 16:46

1 Answer 1


I do not really see your problem, as everything you stated is basically correct and there is no real contradiction as far as I see. Still, I would like to guess what you might be stumbling over.

It is true that the Dirac operators are "the same" in some sense, as they can both be written as $i d/dt$. However, they are not, as they act on different bundles. When writing the operator this way, one makes various identifications, as I explain below. You are right with your statement that for both spin structures, the spinor space are isomorphic to a different bundle. However, the isomorphisms differ.

Let us write $S^1 = \mathrm{R}/\mathrm{Z}$. For the disconnected-cover-spin structure, the associated spinor space then $$ \Sigma_1 = S^1 \times \mathbb{C},$$ a trivial bundle. One can write this as $$ \Sigma_1 = \mathbb{R} \times \mathbb{C} / \sim $$ where $\sim$ is the equivalence relation $$ (t, z) \sim (t^\prime, z^\prime) \Longleftrightarrow t - t^\prime \in \mathbb{Z} ~~ \text{and} ~~ z = z^\prime. $$ For the connected-cover-spin structure, the Spinor space is $$ \Sigma_2 = \mathbb{R} \times \mathbb{C} / \sim$$ where $\sim$ is the equivalence relation $$ (t, z) \sim (t^\prime, z^\prime) \Longleftrightarrow t-t^\prime \in \mathbb{Z} ~~\text{and}~~ z = e^{i \pi (t-t^\prime)} z^\prime $$ This bundle is trivializable but not trivial itself. Now the operator $D = i d/dt$ is usually defined on the space $\mathbb{R} \times \mathbb{C}$, and of course it descends well to operators $D_1$ and $D_2$ on the factor spaces $\Sigma_1$ and $\Sigma_2$. However, $D_1$ and $D_2$ are different operators, and they have different properties; for example, as you mentioned, the kernel of $D_2$ is trivial while the kernel of $D_1$ is not.

\Edit: As any differential operator, the Dirac operator is a local object. In fact, it is given by $$ \sum_{j=1}^n e_j \cdot \nabla_{e_j} $$ for an orthonormal basis $e_1, \dots e_j$, where $\nabla$ is the Levi-Civita connection on Spinors. Now in the case of $S^1$, the manifold is flat, so the Levi-Civita connection coincides with $d$. The spinor bundle is isomorphic to $\mathbb{C}$, by identifying the vector $e_1$ with $i$ (note that there is only one positively oriented local ONB $e_1$ on $S^1$ -- which btw is even globally defined).

This is why in general, Dirac operators always have to coincide locally, in some sense. And also, this is the reason why the Dirac operator on $S^1$ is locally given by $i d/dt$ in any case.

  • $\begingroup$ Maybe this would help me. Why, in the case of the connected spin structure, is the Dirac operator identifiable with $i\frac{d}{dt}$? Another thing that might help: Once we have trivialized the spinor bundle for the connected spin structure, what does the Dirac operator become? $\endgroup$
    – mck
    Commented Dec 1, 2012 at 20:16
  • $\begingroup$ Also see my second edit. Thank you for helping me figure this out. $\endgroup$
    – mck
    Commented Dec 1, 2012 at 21:15

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