2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $2$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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