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 $M_0^{ij} = 0$.

  • $\begingroup$ 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! $\endgroup$ – Pedro Dec 25 '09 at 12:09
  • $\begingroup$ 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 $(B^i \wedge B^j)_0$. $\endgroup$ – José Figueroa-O'Farrill Dec 25 '09 at 12:33
  • $\begingroup$ I think that when one says traceless it is not necessarily "all diagonal element vanishing"... $\endgroup$ – Pedro Dec 25 '09 at 13:15
  • $\begingroup$ 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 $\endgroup$ – Pedro Dec 25 '09 at 13:20
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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?) $\endgroup$ – Pedro Dec 24 '09 at 16:19
  • $\begingroup$ what if $X$ and $Y$ are skew-symmetric? $\endgroup$ – Pedro Dec 24 '09 at 18:14
  • $\begingroup$ 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. $\endgroup$ – j.c. Dec 24 '09 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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