Return to Question

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?

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?

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?

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?