Inverse of an operator does not exist, can't see why

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Elwin.Martin
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I feel kind of lame, but here's my situation:
We start with the operator [itex]g_{\mu \nu} \Box - \partial_{\mu}\partial_{\nu}[/itex] and convert to momentum space to get [itex]-g_{\mu \nu} k^{2} - k_{\mu}k_{\nu}[/itex].

Apparently it's easy to see that this has no inverse?

I'm told that if it *did* it would be of the form
[itex]Ag^{\nu \lambda} +B k^{\nu}k^{\lambda}[/itex]
but I don't see why it would be in this form, to start with. Are these A's and B's just constant coefficients...?

I understand that given the form above, we can simply multiply our inverse and original operator and we get that [itex]-Ak^{2}\delta_{\mu}^{\lambda} +A k_{\mu}k^{\lambda}=\delta_{\mu}^{\lambda}[/itex]
. . . but I don't see why this is an issue, or rather, I can't see why this equation has no solution.

Thanks for any and all help,
E_Martin
 
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Mentz114 said:
I don't think these equations have a solution

[tex] -Ak^{2}\delta_{\mu}^{\lambda} +A k_{\mu}k^{\lambda}=\delta_{\mu}^{\lambda}[/tex]

but I can't prove it right now.

:P then you're in agreement with Ryder...I don't see it though. I have the feeling it doesn't, but I cannot show it.
 
Elwin.Martin said:
:P then you're in agreement with Ryder...I don't see it though. I have the feeling it doesn't, but I cannot show it.

Sorry, I deleted my post because I'm looking for a solution and it might be on.

Have you found a solution ?
[Edit] Any solution leads to a contradiction, so there isn't one. Try writing out all 16 equations.
 
Last edited:
Mentz114 said:
Sorry, I deleted my post because I'm looking for a solution and it might be on.

Have you found a solution ?
[Edit] Any solution leads to a contradiction, so there isn't one. Try writing out all 16 equations.

OH, WOW...thanks!

Just writing out the first component works, haha...fail.

Thanks again!
 
Non-trivial projection operator has no inverse; [itex]P^{2}=P[/itex]. Consider
[tex]P_{ab} = g_{ab} - \frac{k_{a}k_{b}}{k^{2}},[/tex]
[tex]P^{ac}= g^{ac} - \frac{k^{a}k^{c}}{k^{2}}.[/tex]
then
[tex]P_{ab}P^{ac} = \delta^{c}_{b} - \frac{k_{b}k^{c}}{k^{2}} \equiv P^{c}_{b}[/tex]