# 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?

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 $M_0^{ij} = 0$.

• I was wrong about $B^i \wedge B^j$ sorry. "Since $\Phi_{ij}$ is traceless, only the traceless part of $B^i \wedge B^j$ that couples to $\Phi$": I thought about that, In the same way as an antisymmetric tensor kills the symmetric part of a tensor contracted with it... But when I tried this on a concrete example it didn't work! – Pedro Dec 25 '09 at 12:09
• What concrete example? If you write $B^i \wedge B^j$ in the action as $(B^i \wedge B^j)_0$, say, which is traceless, and a trace part, then the third term does not see the trace part. hence the equation of motion of $\Phi$ cannot possibly set all of $B^i \wedge B^j$ to zero, only $(B^i \wedge B^j)_0$. – José Figueroa-O'Farrill Dec 25 '09 at 12:33
• I think that when one says traceless it is not necessarily "all diagonal element vanishing"... – Pedro Dec 25 '09 at 13:15
• if all diagonal elements of $\Phi$ are zero it's ok, one keep only the traceless part of $B^i \wedge B^j$... Perhaps what all these authors wanted to say by traceless is "all diagonal elements" :-s – Pedro Dec 25 '09 at 13:20
• Traceless does not of course mean "all diagonal elements", just their sum. Take a symmetric $N \times N$ matrix $M_{ij}$ (in your case valued in 2-forms...) and decompose it into traceless and trace as follows: $M_{ij} = (M_{ij} - \frac{1}{N} \delta_{ij} t) + \frac{1}{N} \delta_{ij} t$, with $t$ the trace. Then contracting with $\Phi$, you find that because $\Phi$ is traceless, meaning $\Phi_{ij} \delta_{ij} = 0$ (Einstein summation), the trace part gives zero. – José Figueroa-O'Farrill Dec 25 '09 at 13:48

[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.

• Yep; seen like that you're right... The problem is that in all papers I read it is my first answer that appears!! (of course we have a real action ie: a functional, does it change something?) – Pedro Dec 24 '09 at 16:19
• what if $X$ and $Y$ are skew-symmetric? – Pedro Dec 24 '09 at 18:14
• Well if they are skew-symmetric, they are automatically traceless since their diagonal entries must be zero. I suspect that things may become a little clearer if you update your question with a more realistic functional that you're interested in. – j.c. Dec 24 '09 at 18:36