Extremum under variations of a traceless matrix 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?
 A: [MOD: this is an answer to a previous version of the question]
I'm not sure I believe your answer.  Perhaps I'm missing something though.
Let $T = X_{ij}Y^{ij} + \lambda (X_{ij}\delta^{ij})$, which is your original function plus a Lagrange multiplier for the traceless constraint.
Extremize by setting partial derivatives with respect to the entries $X_{ij}$ to zero:
$0=\frac{\partial T}{\partial X_{ij}}=Y^{ij}+\lambda \delta^{ij}$
For entries where $i=j$, this is $Y^{ii}+\lambda =0$, which yields the condition that all diagonal entries of $Y$ are equal, not that $Y$ is traceless.  For the entries with $i\neq j$, we recover $Y^{ij}=0$ as usual.
A: 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.

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 $M_0^{ij} = 0$.
