Before stating the claim, let my define all the objects which I need: To start with, let me fix notation:

  • Let $P$ be a principal $G$-bundle over a (smooth, compact, oriented) manifold $\mathcal{M}$ (possibly with boundary) and $\mathfrak{g}$ be the Lie algebra of $G$. Furthemore, let us choose a connection $1$-form $A\in\Omega^{1}(P,\mathfrak{g})$.
  • Let $(V,\rho)$ be a (finite-dimensional real/complex) representation of $G$ and $E:=P\times_{\rho} V$ be the associated vector bundle.
  • Let $\langle\cdot,\cdot\rangle_{V}$ be a non-degenerate symmetric bilinear form on $V$. Then it is a general fact that this induces a bundle metric $\langle\cdot,\cdot\rangle_{E}\in\Gamma(E^{\ast}\otimes E^{\ast})$ on $E$ via $$\langle [p,v],[p,w]\rangle_{E_{x}}:=\langle v,w\rangle_{V}$$ for all $x\in\mathcal{M}$ and for all $[p,v],[p,w]\in E_{x}\cong P_{x}\times_{\rho}V$.

Furthermore, I need two further definitions: First of all, the wedge-product $\mathrm{tr}(\cdot\wedge\cdot):\Omega^{k}(\mathcal{M},E)\times\Omega^{l}(\mathcal{M},E)\to\Omega^{k+l}(\mathcal{M})$ is defined in the obvious way, i.e. $$\mathrm{tr}(\alpha\wedge\beta)_{x}(v_{1},\dots,v_{k+l}):=\frac{1}{k!l!}\sum_{\sigma\in\mathfrak{S}^{k+l}}\mathrm{sgn}(\sigma)\langle \alpha_{x}(v_{\sigma(1)},\dots,v_{\sigma(k)}),\beta_{x}(v_{\sigma(k+1)},\dots,v_{\sigma(k+l)})\rangle_{E_{x}}$$ for all $x\in\mathcal{M}$ and for all $v_{1},\dots,v_{k+l}\in T_{x}\mathcal{M}$. Secondly, I need the "exterior covariant derivative induced by a connection 1-form $A$", which is the exterior covariant derivative induced by a connection $\nabla^{A}$ on $E$, whose definition is not so important right now. This is a map $\mathrm{d}_{A}:\Omega^{k}(\mathcal{M},E)\to\Omega^{k+1}(\mathcal{M},E)$ defined using a local frame $\{e_{a}\}_{a}\subset\Gamma(U,E)$ defined on some open set $U\subset\mathcal{M}$ via $$\mathrm{d}_{A}\alpha\vert_{U}:=\sum_{a}(\mathrm{d}\alpha^{a}e_{a}+(-1)^{k}\alpha^{a}\wedge\nabla^{A}e_{a})$$ where $\alpha\vert_{U}=\sum_{a}\alpha^{a}e_{a}$ for coordinates $\alpha^{a}\in\Omega^{k}(U)$. (Strictly speaking, I should write $\alpha^{a}\otimes e_{a}$, but let me keep notation simple)

Now I would like to prove the following:


I have to say that I am not sure if this is actually true in this form. It is rather an educated guess. For context, the reason for this is that something like this is implicitely used on page $4$ of arXiv:gr-qc/9905087 in the proof of invariance of the BF-action under translational symmetry. (In this context, the bundle $E$ is given by the adjoint bundle $\mathrm{Ad}(P)$). In this paper, the authors used "integration by parts" for an expression of the type $\mathrm{tr}(\mathrm{d}_{A}\eta\wedge F)$, where $\eta\in\Omega^{d-3}(\mathcal{M},\mathrm{Ad}(P))$ and where $F\in\Omega^{2}(\mathcal{M},\mathrm{Ad}(P))$ denotes the curvature of $A$.

Now my attempt is the following: I think it is easier to proof this in the local frame on $U$. Let us write $\alpha\in\Omega^{k}(\mathcal{M},E)$ and $\beta\in\Omega^{l}(\mathcal{M},E)$ in this frame, i.e. $$\alpha\vert_{U}=\sum_{a}\alpha^{a}e_{a}\hspace{1cm}\text{and}\hspace{1cm}\beta\vert_{U}=\sum_{a}\beta^{a}e_{a}$$ for real-valued coordinate forms $\alpha^{a}\in\Omega^{k}(U)$,$\beta^{a}\in\Omega^{l}(U)$. Then the above defined trace-wedge product is in coordinates given by $$\mathrm{tr}(\alpha\wedge\beta)\vert_{U}=\sum_{a,b}(\alpha^{a}\wedge\beta^{b})\langle e_{a},e_{b}\rangle_{E}$$

where $\langle e_{a},e_{b}\rangle_{E}$ is defined in the obious way, i.e. $\langle e_{a},e_{b}\rangle_{E}(x):=\langle e_{a}(x),e_{a}(x)\rangle_{E_{x}}$. With this, the left-hand side of the conjectured equation is given by $$\mathrm{d}(\mathrm{tr}(\alpha\wedge\beta))\vert_{U}=\sum_{a,b}(\mathrm{d}\alpha^{a}\wedge\beta^{b}+(-1)^{k}\alpha^{a}\wedge\mathrm{d}\beta^{b})\langle e_{a},e_{b}\rangle_{E}$$ where we just used the standard Leibniz rule for real-valued forms. Now for the right-hand side, let us firstly write $\nabla^{A}e_{a}$ in terms of local connection $1$-forms ${\omega^{i}}_{j}\in\Omega^{1}(U)$ via $$\nabla^{A}e_{a}=\sum_{b}{\omega^{b}}_{a}e_{b}.$$ With this, we have that


Now we know how the coordinate forms of $\mathrm{d}_{A}\alpha$ look like and hence we can compute the right-hand side: First of all, we have that

$$\mathrm{tr}(\mathrm{d}_{A}\alpha\wedge\beta)\vert_{U}=\sum_{a,b}(\mathrm{d}\alpha^{a}\wedge\beta)\langle e_{a},e_{b}\rangle_{E}+(-1)^{k}\sum_{a,b,c}(\alpha^{c}\wedge{\omega^{a}}_{c}\wedge\beta^{b})\langle e_{a},e_{b}\rangle_{E}$$

Completely analogues, we find that

$$\mathrm{tr}(\alpha\wedge\mathrm{d}_{A}\beta)\vert_{U}=\sum_{a,b}(\alpha^{a}\wedge\mathrm{d}\beta^{b})\langle e_{a},e_{b}\rangle_{E}+(-1)^{l}\sum_{a,b,c}(\alpha^{a}\wedge\beta^{c}\wedge{\omega^{b}}_{c})\langle e_{a},e_{b}\rangle_{E}$$

Hence, if the above formula is true, we must have that

$$\sum_{a,b,c}(\alpha^{c}\wedge{\omega^{a}}_{c}\wedge\beta^{b}+(-1)^{l}\alpha^{a}\wedge\beta^{c}\wedge{\omega^{b}}_{c})\langle e_{a},e_{b}\rangle_{E}\stackrel{!}{=}0$$

I can't see why this is the case. Again, I should stress that I am not even sure if the claimed equality is true. Maybe it is false, or maybe the factor $(-1)^{k}$ is different. Furthemore, in the paper cited above this is only used for the case $E=\mathrm{Ad}(P)$, $k=d-2$ and $l=2$, where $d=\mathrm{dim}(\mathcal{M})$, so maybe it is only true in this specific case. Any help and comment is appreciated!

  • 1
    $\begingroup$ In your expression for $d \left( \operatorname{tr} \left( \alpha \wedge \beta \right) \right)$ you forgot to take into account that $\left< e_a, e_b \right>$ is a function so you need to take its differential. Then you get an extra term of $\sum_{a,b} d \left< e_a, e_b \right> \wedge \alpha^a \wedge \beta^b$. I haven't checked all the details but I'm pretty sure that if you assume that $\nabla$ is metric with respect to $\left< \cdot, \cdot \right>$ so that $d \left< e_a, e_b \right> = \left< \nabla e_a, e_b \right> + \left< e_a, \nabla e_b \right>$ then you're desired formula $\endgroup$
    – levap
    Aug 2, 2021 at 13:01
  • 1
    $\begingroup$ will be true. When $E = \operatorname{Ad}(P)$ and the inner product comes from an symmetric invariant form on $\mathfrak{g}$ this will automatically hold. $\endgroup$
    – levap
    Aug 2, 2021 at 13:02
  • $\begingroup$ Thank you very much @levap. Indeed, I completely forgot about it! $\endgroup$ Aug 2, 2021 at 13:43
  • $\begingroup$ And yes indeed, the covariant derivative $\nabla^{A} $ induced by the connection 1-form $A$ is compatible with the bundle metric for all associated vector bundles, i.e. it fulfills the metric condition $L_{X} \langle s, t\rangle_{E} = \langle \nabla^{A}_{X} s, t\rangle_{E} + \langle s, \nabla^{A}_{X} t\rangle_{E}$ for all vector fields $X$ and for all sections $s, t$ of $E$, whenever we take the bundle metric $\langle \cdot, \cdot\rangle_{E} $ induced by a $V$-invariant inner product. $\endgroup$ Aug 2, 2021 at 13:50

2 Answers 2


Let me restate everything in the abstract index notation, preserving some of the notation from the question, but adapting where it makes more sense to me. In particular, I try to keep the indices balanced.

The definition of the wedge product can be restated as $$ (\alpha^\Phi \wedge \beta^\Psi)_{[i_{1} \dots i_{k} i_{k + 1} \dots i_{k + l} ]} = \alpha^\Phi_{[i_{1} \dots i_{k}} \beta^\Psi_{i_{k + 1} \dots i_{k + l} ]} $$

Similarly, the definition of the exterior derivative has an equivalent form as follows $$ (\mathrm{d}^A \alpha^\Phi)_{[i_{0} i_{1} \dots i_{k}]} = \nabla^A_{[i_{0}} \alpha^\Phi_{i_{1} \dots i_{k}]} $$

Clearly, the identity $$ \mathrm{d}^A (\alpha^\Phi \wedge \beta^\Psi) = (\mathrm{d}^A \alpha^\Phi) \wedge \beta^\Psi + (-1)^k \alpha^\Phi \wedge \mathrm{d}^A \beta^\Psi $$ is purely algebraic and its proof is well known.

However, the question is about the "trace". In other words, it is assumed additionally, that there is "an invariant nondegenerate bilinear form" on the Lie algebra $G$ in the original structure, but because a representation is fixed (via the adjoint action of $G$), we work entirely with the associated vector bundles. Let us denote the section, corresponding to this bilinear form as $h_{\Phi \Psi}$. This is a bundle metric, which is used to take traces.

The formula for the trace can be presented as $$ \mathrm{tr}(\alpha^\Phi \wedge \beta^\Psi) = h_{\Phi \Psi} \alpha^\Phi \wedge \beta^\Psi $$

With this restatements, we can rephrase the identity in question: $$ \mathrm{d}^A (h_{\Phi \Psi} \alpha^\Phi \wedge \beta^\Psi) = h_{\Phi \Psi} (\mathrm{d}^A \alpha^\Phi) \wedge \beta^\Psi + (-1)^k h_{\Phi \Psi} \alpha^\Phi \wedge \mathrm{d}^A \beta^\Psi $$ but it would only hold universally, if $$ \nabla^A h_{\Phi \Psi} \stackrel{!}{=} 0 $$ that is, in the case when the associated connection is metric with respect to the chosen bilinear form.

This last equation looks suspiciously equivalent to the last equation in the question. I am not too sure, if it is true or not, because I am not quite familiar with the properties of the vector bundles, associated via the adjoint action, but I have the gut feeling that the bottom line is somewhere there.

  • $\begingroup$ Thank you very much for the answer. Indeed, it is related to the metric compatibility of the connection. I am not so familiar with the abstract index notation, but it seems to be very helpful, so maybe I should have a look at it... $\endgroup$ Aug 2, 2021 at 17:15

For completeness, let me post an answer to my question. First of all, the claim I wanted to proof is correct. As pointed out in the comment by @levap, I forgot to take the derivative of $\langle e_{a},e_{b}\rangle_{E}\in C^{\infty}(U)$ on the left-hand side. It follows that the claim is related to the metric compatibility of the induced bundle metric, as already pointed out in the answer by @Yuri Vyatkin. Thank you very much for all the helpful comments!

First of all, the correct expression for the left-hand side reads

$$\mathrm{d}(\mathrm{tr}(\alpha\wedge\beta))\vert_{U}=\sum_{a,b=1}^{\mathrm{dim}_{\mathbb{R}}(V)}(\mathrm{d}\alpha^{a}\wedge\beta^{b}+(-1)^{k}\mathrm{d}\alpha^{a}\wedge\mathrm{d}\beta^{b})\langle e_{a},e_{b}\rangle_{E}+(-1)^{k+l}\sum_{a,b=1}^{\mathrm{dim}_{\mathbb{R}}(V)}\alpha^{a}\wedge\beta^{b}\wedge\mathrm{d}\langle e_{a},e_{b}\rangle_{E}.$$

The right-hand side is given by

$$(\mathrm{tr}(\mathrm{d}_{A}\alpha\wedge\beta)+(-1)^{k}\mathrm{tr}(\alpha\wedge\mathrm{d}_{A}\beta))\vert_{U}=$$ $$=\sum_{a,b=1}^{\mathrm{dim}_{\mathbb{R}}(V)}(\mathrm{d}\alpha^{a}\wedge\beta)\langle e_{a},e_{b}\rangle_{E}+(-1)^{k}\sum_{a,b,c=1}^{\mathrm{dim}_{\mathbb{R}}(V)}(\alpha^{c}\wedge{\omega^{a}}_{c}\wedge\beta^{b})\langle e_{a},e_{b}\rangle_{E}+$$ $$+(-1)^{k}\sum_{a,b=1}^{\mathrm{dim}_{\mathbb{R}}(V)}(\alpha^{a}\wedge\mathrm{d}\beta^{b})\langle e_{a},e_{b}\rangle_{E}+(-1)^{k+l}\sum_{a,b,c=1}^{\mathrm{dim}_{\mathbb{R}}(V)}(\alpha^{a}\wedge\beta^{c}\wedge{\omega^{b}}_{c})\langle e_{a},e_{b}\rangle_{E}$$

As a consequence, the claim I wanted to prove is true if and only if

$$\sum_{a,b=1}^{\mathrm{dim}_{\mathbb{R}}(V)}\alpha^{a}\wedge\beta^{b}\wedge\mathrm{d}\langle e_{a},e_{b}\rangle_{E}\stackrel{!}{=}\sum_{a,b,c=1}^{\mathrm{dim}_{\mathbb{R}}(V)}(\alpha^{c}\wedge\beta^{b}\wedge{\omega^{a}}_{c}+\alpha^{a}\wedge\beta^{c}\wedge{\omega^{b}}_{c}))\langle e_{a},e_{b}\rangle_{E}$$

Now as written in my comment, the connection $\nabla^{A}$ is metric with respect to the induced bundle metric $\langle\cdot,\cdot\rangle_{E}$ thanks to $V$-invariance of $\langle\cdot,\cdot\rangle_{V}$, i.e.

$$\mathrm{d}(\langle e_{a},e_{b}\rangle_{E})=\langle\nabla^{A}e_{a},e_{b}\rangle_{E}+\langle e_{a},\nabla^{A}e_{b}\rangle_{E}$$

Using the local connection $1$-forms, this can be written as

$$\mathrm{d}(\langle e_{a},e_{b}\rangle_{E})=\sum_{c=1}^{\mathrm{dim}_{\mathbb{R}}(V)}{\omega^{c}}_{a}\langle e_{c},e_{b}\rangle_{E}+\sum_{c=1}^{\mathrm{dim}_{\mathbb{R}}(V)}{\omega^{c}}_{b}\langle e_{a},e_{c}\rangle_{E}$$

and hence, we exactly get that

$$\sum_{a,b=1}^{\mathrm{dim}_{\mathbb{R}}(V)}\alpha^{a}\wedge\beta^{b}\wedge\mathrm{d}\langle e_{a},e_{b}\rangle_{E}=\sum_{a,b,c=1}^{\mathrm{dim}_{\mathbb{R}}(V)}(\alpha^{c}\wedge\beta^{b}\wedge{\omega^{a}}_{c}+\alpha^{a}\wedge\beta^{c}\wedge{\omega^{b}}_{c})\langle e_{a},e_{b}\rangle_{E}$$

as claimed.


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