The gauge-invariant part in Yang-Mills Lagrangian is $$ \mathcal{L}_{\text{gauge}} = -\frac{1}{2}TrF_{\mu\nu}F^{\mu\nu} = -\frac{1}{4}F_{\mu\nu}^aF^{a, \mu\nu}. $$ Sometimes I see the lagrangian written as $$ \mathcal{L}_{\text{gauge}} = \frac{1}{2}Tr(F∧⋆F). $$ I have only seen this notation when I learned the wedge product of two vectors could be visualized as the area of a parallelogram, and the Hodge $⋆$ operator gives the perpendicular vector which makes the product positively oriented. How do I understand the Hodge operator and wedge product here?


2 Answers 2


The Hodge dual of a $p$-form (antisymmetric $(0,p)$ tensor) in $d$ dimensions is a $(d-p)$-form whose components are defined by $$ (\star A)_{\mu_1 \cdots \mu_{d-p}} \equiv \frac{1}{p!} \epsilon_{\mu_1 \cdots \mu_{d-p} \nu_1 \cdots \nu_p} A^{\nu_1 \cdots \nu_p} . $$ The normalization $1/p!$ is a user-dependent convention so different books use different choices. $\epsilon$ is the volume $d$-form. It is a completely antisymmetric $(0,d)$ tensor with $$ \epsilon_{01\cdots(d-1)} = \sqrt{|g|} . $$ This along with complete antisymmetry uniquely specifies $\epsilon$.

The wedge product $\wedge$ is a exterior product on forms. The wedge product of a $p$-form and a $q$-form is a $p+q$ form whose components are given by $$ ( A \wedge B )_{\mu_1 \cdots \mu_{p+q} } \equiv \frac{(p+q)!}{p!q!} A_{[\mu_1 \cdots \mu_p} B_{\mu_{p+1} \cdots \mu_{p+q} ] } . $$ where the square brackets $[\cdots]$ denotes the weighted antisymmetrization.

Finally, we note that up to a normalization, there is a unique $d$-form, namely the volume form. So any $d$ form $X$ can be written as $$ X = \left( - \frac{1}{d!} \epsilon^{\mu_1 \cdots \mu_d} X_{\mu_1 \cdots \mu_d} \right) \epsilon . $$ The extra minus sign above comes from the assumption that the metric is Lorentzian.

We are now ready to process the term in the question. In this case, $F$ is a two form so $\star F$ is a $(d-2)$-form. $F \wedge \star F$ is a $d$-form. Using the definitions I've given you above, we find \begin{align} F \wedge \star F &= \left( - \frac{1}{d!} \epsilon^{\mu_1 \cdots \mu_d} [ F \wedge \star F ]_{\mu_1 \cdots \mu_d} \right) \epsilon \\ &= \left( - \frac{1}{d!} \frac{d!}{(d-2)!2!} \epsilon^{\mu_1 \cdots \mu_d} F_{\mu_1 \mu_2 } ( \star F )_{\mu_3 \cdots \mu_d} \right) \epsilon \\ &= \frac{1}{2}\left( - \frac{1}{(d-2)!2!} \epsilon^{\mu_1 \cdots \mu_d} \epsilon_{\mu_3 \cdots \mu_d \nu_1 \nu_2} F_{\mu_1 \mu_2 } F^{\nu_1 \nu_2} \right) \epsilon \end{align} Next, we use the following property that the volume form satisfies $$ \epsilon^{\mu_1 \cdots \mu_d} \epsilon_{\mu_3 \cdots \mu_d \nu_1 \nu_2} = - (d-2)! [ \delta^{\mu_1}_{\nu_1} \delta^{\mu_2}_{\nu_2} - \delta^{\mu_2}_{\nu_1} \delta^{\mu_1}_{\nu_2} ] . $$ Then, \begin{align} F \wedge \star F &= \left( \frac{1}{2} F_{\mu\nu} F^{\mu\nu} \right) \epsilon . \end{align} Including the extra factor of $-\frac{1}{2}$ and the Lie algebra trace (which plays no role in these manipulations), we find \begin{align} - \frac{1}{2} \text{Tr}[F \wedge \star F] &= \left( - \frac{1}{4} \text{Tr} [ F_{\mu\nu} F^{\mu\nu} ] \right) \epsilon . \end{align} Integrating both sides, we then find \begin{align} - \frac{1}{2} \int \text{Tr}[F \wedge \star F] &= \int \textrm{d}^d x \left( - \frac{1}{4} \text{Tr} [ F_{\mu\nu} F^{\mu\nu} ] \right) . \end{align}


Intuitively, you can think of the Hodge dual and outer product geometrically as you already do, and this combination as a sort of generalised dot product.

$A\wedge B$ has a magnitude that extracts the component of $A$ perpendicular to $B$, times the length of $B$. So $A\wedge \star B$ has a magnitude that extracts the component of $A$ perpendicular to $\star B$ and hence parallel to $B$, times the length of $B$. For vectors (or 1-forms), this is just the dot product $A\cdot B$. Because we already know how to take the outer product of higher order forms, we don't have to develop dot products from scratch. This allows us to define dot products of higher order forms in terms of operations we have already defined.

$F\wedge \star F$ is thus the dot product of $F$ with itself, which is its magnitude squared. Because we are taking the outer product of $F$ with its orthogonal complement $\star F$, the result is the fixed unit volume form covering all the dimensions, multiplied by the dot product. The trace operation is then used to turn this volume form into a scalar.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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