Let $Y$ be a closed oriented $2$-dimensional manifold, $G$ a Lie group and $Q \to Y$ a principal $G$-bundle with a given section $q.$ Denote by $\mathcal{A}_Q$ the space of connections on $Q,$ and by $L_Q \to \mathcal{A}_Q$ the Chern-Simons line bundle. Suppose we have an Ad-invariant symmetric bilinear form $$\langle -,- \rangle: \mathfrak{g} \times \mathfrak{g} \to \mathbb{C}.$$

Given this data, we can then define the $1$-form $\theta_q$ on $\mathcal{A}_Q$ by $$(\theta_q)_\eta(\dot{\eta}) = 2\pi i \int_Y q^*\langle \eta \wedge \dot{\eta} \rangle, \: \: \:\eta \in \mathcal{A}_Q, \: \dot{\eta} \in T_\eta \mathcal{A}_Q.$$

I have seen (for example, in the paper Classical Chern-Simons Theory, Part I by Freed, pg. 27) the claim that $$\dfrac{i}{2\pi} d(\theta_q)(\dot{\eta_1},\dot{\eta_2}) = -2 \int_Y q^*\langle \dot{\eta_1},\dot{\eta_2} \rangle.$$ Here $d$ denotes the exterior derivative. Is this true? I do not see where the factor of two arises from. I would expect this to follow if we could show that $$d\langle \eta,\dot{\eta} \rangle(\dot{\eta_1},\dot{\eta_2}) = 2 \langle \dot{\eta_1},\dot{\eta_2} \rangle.$$ However, this is not what I get. I would expect the exterior derivative to act as $$d\langle \eta,\dot{\eta} \rangle = \langle d(\eta),\dot{\eta} \rangle + \langle \eta,d(\dot{\eta}) \rangle.$$ Now, further, I would think that $d(\eta) = \dot{\eta},$ so that since $d^2=0,$ I get, when evaluating this on $(\dot{\eta_1},\dot{\eta_2})$ just $\langle \dot{\eta_1},\dot{\eta_2} \rangle.$

So I would like to ask:
Is the claimed equality true? If so, why is it true? What is wrong with my proposed way of going about calculating it?


For simplicity, suppose $V$ is just a vector space (finite-dimensional, if you like). Let $\omega\in\Lambda^2 V^\vee$ (you start with a symmetric bilinear form, but combining it with the antisymmetric integral pairing of one-form gives an antisymmetric bilinear form). Then there are two forms associated to it: the constant two-form $\omega$ and the one-form $\xi_\omega$ whose value on a tangent vector $Y\in T_XV$ is $\omega(X,Y)$. Here $X\in V$ is regarded as a point of the manifold $V$, and $Y$ is regarded as a (constant) vector field on $V$.

Now let $X,Y\in V$ and regard them as (constant) vector fields on $V$. Then the usual formula for the differential gives \begin{align*} \mathrm d\xi_\omega(X,Y) &= X(\xi_\omega(Y)) - Y(\xi_\omega(X)) - \xi_\omega([X,Y])\\ &= \omega(X,Y) - \omega(Y,X) - 0\\ &= 2\omega(X,Y) \end{align*}

Alternatively, consider the ``Euler'' vector field $E$ whose value at a point $X\in V$ is $X\in T_XV\cong V$. Then by definition, $\xi_\omega = \iota_E\omega$. Since $\omega$ is closed, Cartan's formula gives $\mathrm d\xi_\omega = \mathcal L_E\omega = 2\omega$ since $E$ is the infinitesimal generator of scalar multiplication, so a constant coefficient $k$-form has weight $k$ under it.

More alternatively, suppose we consider some vector space $V$ over a field $K$ of characteristic $2$. Then there is still an exact sequence $(V\otimes V)^{C_2}\to V\otimes V\to (V\otimes V)_{C_2}$ of $GL_K(V)$-modules which essentially plays the role of the deRham complex of $V$, but it no longer splits $GL_K(V)$-equivariantly. This means that any construction which takes a constant coefficient antisymmetric form and produces a primitive for it must divide by $2$ or use an explicit basis (otherwise the formula would still make sense over $K$).

As to your calculation that $\mathrm d\eta = \dot\eta$, the only way I know how to make sense of the left-hand side is to say that $\eta = E$ is the Euler vector field and $\mathrm d$ is the canonical affine connection on the trivial tangent bundle of $V$. Thus $\mathrm d\eta$ should be a one-form with values in the tangent bundle (it is of course the canonical one-form). We do indeed have $\mathrm d\omega(\eta,\dot\eta) = \omega(\dot\eta,\dot\eta)$, but this is not the constant coefficient two-form determined by $\omega$: It takes two tangent vectors and first produces $\dot\eta\wedge\dot\eta(X,Y) = X\otimes Y - Y\otimes X$, then evaluates $\omega$ on it. Since $\omega$ was already antisymmetric, we get the same result twice, which explains the overall factor of $2$.

  • $\begingroup$ Thank you for your answer. I have three questions. First, what do you mean by the constant differential form associated to $\omega?$ Second, when you write $\eta$ in the first equation display, do you mean $\omega?$ Lastly, why is $X\eta(Y)$ equal to $\omega(X,Y)?$ $\endgroup$ – Dedalus Jan 28 at 22:43
  • $\begingroup$ The bundle of two-forms is trvialized, so any element of $\Lambda^2 V^\vee$ determines a two-form on the manifold $V$. If $\omega = \omega_{ij}e^i\wedge e^j$ then the corresponding two-form is $\omega_{ij}\mathrm dx^i\mathrm dx^j$, which has constant coefficients. More rigourously, $\omega$ is flat with respect to the canonical torsion-free flat connection on the affine manifold $V$. $\eta$ is correct, the expression $X(\eta(Y))$ means that the vector field $Y$ is plugged into the one-form $\eta$ to produce a smooth function which can be differentiated along $X$. $\endgroup$ – Bertram Arnold Jan 28 at 23:25
  • $\begingroup$ This function is $\omega(-,Y)$, which is linear, so its derivative along $X$ is $\omega(X,Y)$. $\endgroup$ – Bertram Arnold Jan 28 at 23:26
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    $\begingroup$ Both of these are fine, the point is that (in my notation) $\omega(\dot\eta,\dot\eta)(X,Y) = \omega(X,Y) - \omega(Y,X) = 2\omega(X,Y)$. Thus your second-to-last equation should read $\mathrm d\langle \eta,\dot\eta\rangle = \langle \dot\eta,\dot\eta\rangle$. Fixing a basis $e_i$ of $V$, we have $\eta = e_i\otimes x^i,\dot\eta = e_i\otimes \mathrm dx^i, \omega(\dot\eta,\dot\eta) = \omega(e_i,e_j)\mathrm dx^i\mathrm dx^j$, whereas the 2-form $X\otimes Y\mapsto \omega(X,Y)$ is given by $\frac{1}{2}\omega(e_i,e_j)\mathrm dx^i\mathrm dx^j$. $\endgroup$ – Bertram Arnold Jan 29 at 17:04
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    $\begingroup$ Again, I can recommend Freed's Five lectures on Supersymmetry for examples of these calculations in index and geometric notation. The conceptual point is that 2-forms sometimes arise as bilinear functions vanishing on symmetric tensors and sometimes as functions on antisymmetric tensors. When you pass between the two notions you get a factor of $2$. Where exactly this factor shows up is as much a matter of convention as anything. I have edited my answer since I used $\eta$ twice for different things, if you still don't understand it try to do it with indices to see what's going on. $\endgroup$ – Bertram Arnold Jan 29 at 17:13

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