Sorry for my precedent tentative, I was a little hasty:


Ok, I think I'd better put the original problem:


I have an action of three fields: $A$ which is the spin-connection, $B$ an skew-symmetric 2-form and $\Phi$ which is traceless ~~and skew-symmetric~~ scalar field. These fields take their values on some algebra, index their components in this algebra by $i,j,k,... = 1,2,3$


I want to implement a certain condition on B by using equations of motion of $\Phi$, the action is:


$S=\int (B\_i \wedge F^i + \Lambda B\_i \wedge B^i + \Phi\_{ij} B^i \wedge B^j) $


Now for me equations of motions are simply:


$B^i \wedge B^j=0$


perhaps with the condition that all diagonal elements are equal (as jc showed) ~~but this is automatically satisfied for a skew-symmetric matrix (here $B^i \wedge B^j$)~~.


But in all papers I find:


$B^i \wedge B^j - \frac{1}{3}\delta^{ij}B\_k\wedge B^k = 0$


So I see that they all took the traceless part of the matrix representing equations of motion, necessarily it has a relation with the traceless character of $\Phi$ but I do not see which one.


~~In addition, this expression is not antisymmetric in $i,j$~~.


Would anyone have an idea?



Extremum under variations of a traceless matrix
If $B^i$ are 2-forms, then $B^i \wedge B^j$ is symmetric, not skewsymmetric. Since $\Phi\_{ij}$ is traceless, only the traceless part of $B^i \wedge B^j$ that couples to $\Phi$. So I see nothing wrong with the equation you find in the papers.


The reason you take $\Phi$ to be traceless is that the trace is already contained in the second term in the action.




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**Edit** (in response to the comment thread below)


Let me give more details. Let $M^{ij} := B^i \wedge B^j$. We can think of $M$ as a matrix with 4-forms as entries. The space of even forms is a commutative algebra, so we can work with $M$ as if it were a real or complex matrix, say. In particular, we can take its trace (which will be a 4-form): $T = \delta\_{ij} M^{ij}$, where as in the question the Einstein summation convention is in force. We can then decompose $M$ into a traceless part we shall call $M\_0$ and a part containing the trace:
$$M^{ij} = M\_0^{ij} + \frac{1}{N} T \delta^{ij},$$
where I assume that $M$ is an $N\times N$ matrix. If you take the trace of this equation, you find that $M\_0$ is indeed traceless. Its explicit form is given by solving that equation for $M\_0$, but we do not need it.


Now let $\Phi\_{ij}$ be a symmetric traceless matrix. This means that $\delta^{ij} T\_{ij} = 0$. Contracting with $M$ we find
$$\Phi\_{ij} M^{ij} = \Phi\_{ij} M\_0^{ij}.$$
In other words, $\Phi$ never sees the trace of $M$ and hence if you have a lagrange multiplier term in an action functional of the form
$$\int \Phi\_{ij} M^{ij}$$
this is really equal to
$$\int \Phi\_{ij} M\_0^{ij}$$
and hence the resulting Euler-Lagrange equation is $M\_0^{ij} = 0$.


