A question about an identity in deriving Born-Infeld action

I have a question in David Tong's Example Sheet 4 Problem 5b, how to verify the last equation (*) on p.2? (There is a solution for example sheet 3, but seems to be no solution for example sheet 4.)

Problem 5b:

Show that the equations of motion arising from the Born-Infeld action are equivalent to the beta function condition for the open string,
$$\beta_\sigma\left(F\right)=\left( \frac{1} {1 -F^2} \right)^{\mu \rho}\partial_\mu F_{\rho\sigma }=0$$ Note: To do this, it will prove very useful if you can ﬁrst show the following results:
$$∂_μ\left[\operatorname{tr} \ln(1 − F^2)\right] = −4 ∂_\rho F_{μ\sigma}\left(\frac{F}{1-F^2}\right)^{\sigma \rho }$$

which requires use of the Bianchi identity for $F_{\mu \nu }$ and

\tag{*} \begin{align} \partial _\mu \left( \frac{ F}{1-F^2} \right)^{\mu\nu} &= \left( \frac{ F}{1-F^2} \right)^{\mu\rho} \partial_\mu F_{\rho\sigma} \left( \frac{ F}{1-F^2} \right)^{\sigma \nu} \\ &\qquad+ \left( \frac{1} {1 -F^2} \right)^{\mu \rho} \partial_\mu F_{\rho\sigma}\left( \frac{ 1 }{1-F^2} \right)^{\sigma \nu} \end{align}

In addition, as given in question 5a

$$F_{\mu\nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$$

My attempt to prove the problem:

LHS $$\partial_{\mu} \left( \frac{ F}{1-F^2} \right)^{\mu\nu} = \partial_{\mu} \left[ F^{\mu}_{\alpha} \left( \frac{1}{1-F^2} \right)^{\alpha \nu} \right] = \left( \partial_{\mu} F^{\mu}_{\alpha} \right) \left( \frac{1}{1 -F^2} \right)^{\alpha \nu} + F^{\mu}_{\alpha} \partial_{\mu} \left( \frac{1}{1 -F^2} \right)^{\alpha \nu} \tag{1}$$

Using the formula in matrix cookbook for the derivative of inverse matrix, Eq. (53)

Eq. (1) becomes $$\left( \partial_{\mu} F^{\mu}_{\alpha} \right) \left( \frac{1}{1 -F^2} \right)^{\alpha \nu} + 2 F^{\mu}_{\alpha} \left[ \frac{1}{1 -F^2} \left( \partial_{\mu} F \right) F \frac{1}{1-F^2} \right]^{\alpha \nu}$$
$$= \left( \partial_{\mu} F^{\mu}_{\alpha} \right) \left( \frac{1}{1 -F^2} \right)^{\alpha \nu} + 2 \left( \frac{F}{1 -F^2} \right)^{\mu \rho} \left( \partial_{\mu} F \right)_{\rho\sigma} \left(\frac{F}{1-F^2}\right)^{\sigma \nu} \tag{2}$$

The second term in Eq.(2) cancels the second term in the RHS in the problem sheet equation. We then need to show $$\left( \partial_{\mu} F^{\mu}_{\alpha} \right) \left( \frac{1}{1 -F^2} \right)^{\alpha \nu} + \left( \frac{F}{1 -F^2} \right)^{\mu \rho} \left( \partial_{\mu} F \right)_{\rho\sigma} \left(\frac{F}{1-F^2}\right)^{\sigma \nu} - \left( \frac{1}{1 -F^2} \right)^{\mu \rho} \left( \partial_{\mu} F \right)_{\rho\sigma} \left(\frac{1}{1-F^2}\right)^{\sigma \nu} =0 \tag{3}$$

then I didn't find a way to show Eq. (3) hold. I tried to combine the second and third term, and rearrange them, but didn't got a simple expression.

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I think I proved it, I'll write an answer, 1 second... – centralcharge Sep 1 '13 at 2:36
Ok done. (18 minutes, though; instead of 1 second)... Nice "matrix cookbook", by the wayy. – centralcharge Sep 1 '13 at 2:54
Comment to the question (v1): It would be good if OP (or somebody else?) could try to make the question formulation self-contained, so one doesn't have to open the link to understand the question. – Qmechanic Sep 1 '13 at 7:02
@Qmechanic: Ok, done. – centralcharge Sep 1 '13 at 7:31
@DImension10AbhimanyuPS note the inclusion of a crucial omitted factor. – Emilio Pisanty Sep 1 '13 at 11:18

$\partial_i \frac{F}{1-F^2}=-(\frac{F}{1-F^2})\partial_i (\frac{1-F^2}{F})(\frac{F}{1-F^2})$

$=-(\frac{F}{1-F^2})\partial_i(\frac{1}{F}-F)(\frac{F}{1-F^2})$

$=-(\frac{F}{1-F^2})(\partial_i(\frac{1}{F}) -\partial_iF)(\frac{F}{1-F^2})$

$=-(\frac{F}{1-F^2})(- \frac{1}{F}\partial_i F \frac{1}{F}) -\partial_iF)(\frac{F}{1-F^2})$

$=(\frac{1}{1-F^2})\partial_i F(\frac{1}{1-F^2}) + (\frac{F}{1-F^2})\partial_i F(\frac{F}{1-F^2})$

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Eq. ($\ast$) in David Tong's notes follows from the more general eq. ($\ast\ast$)

\tag{**}\begin{align} \partial_{\mu}\left(\frac{F}{1-F^2}\right)^{\lambda}{}_{\nu} &=\left(\frac{F}{1-F^2}\right)^{\lambda}{}_{\rho} ~\partial_{\mu}F^{\rho}{}_{\sigma} \left(\frac{F}{1-F^2}\right)^{\sigma}{}_{\nu} \\ &\qquad+\left(\frac{1}{1-F^2}\right)^{\lambda}{}_{\rho} ~\partial_{\mu} F^{\rho}{}_{\sigma} \left(\frac{1}{1-F^2}\right)^{\sigma}{}_{\nu} \end{align}

by putting the $\lambda$-index equal to the $\mu$-index, sum over $\mu$, and raise the $\nu$-index. Eq. ($\ast\ast$) is equivalent to eq. ($\ast\ast\ast$)

\tag{***}\begin{align} \partial_{\mu}\left(\frac{F}{1-F^2}\right) &=\left(\frac{F}{1-F^2}\right) \partial_{\mu}F \left(\frac{F}{1-F^2}\right) \\ &\qquad+\left(\frac{1}{1-F^2}\right) \partial_{\mu} F \left(\frac{1}{1-F^2}\right), \end{align}

if we implicitly imply that every matrix $F$ carries one upper and one lower index a la $F^{\lambda}{}_{\nu}$. (Note that that there is no ambiguity in writing the matrix expression $\frac{F}{1-F^2}$ as a fraction because the numerator and the denominator commute.)

Finally, eq. ($\ast\ast\ast$) follows by straightforward manipulations, which involve e.g. using the rule

$$\tag{****} \partial_{\mu}\frac{1}{M}~=~-\frac{1}{M}\partial_{\mu}M\frac{1}{M}$$

of how to differentiate an inverse matrix $M$. (In particular, the Bianchi identity that David Tong mentions is not used in the proof of eq. ($\ast\ast\ast$).)

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Thanks for your answer. Excuse me, I tried to use the matrix derivative in my attempt, as Eq. $(****)$ also in the matrix cookbook Eq. (53), and my Eq. (2) is not the same as (****). What is the reason for that? – user26143 Sep 1 '13 at 18:36
It is solved in Trimok's answer, stupid me :( – user26143 Sep 1 '13 at 18:52
Trimok's answer (v2) shows the manipulations that establishes eq. (***). – Qmechanic Sep 1 '13 at 21:26

Non-Rigorous Proof

It's pretty simple algebraic manipulation, it seems. If we let

$$k=\frac{1}{1-F^2}$$

$$n=\partial_\mu F$$

(Letting the indices take care of themselves... You can imagine them "cancelling" out implicitly...)

The statement reduces to:

$$nk+FknFk=knk$$

Which is equivalent to

$$1+F^2k=k$$

Since $k=\frac{1}{1-F^2}$, we see that this is the same as saying that

$$1+\frac{F^2}{1-F^2}=\frac{1}{1-F^2}$$

Which is certainly true!...

Rigorous proof

You can also prove it rigorously as follows:

$$\begin{gathered} {\left( {\frac{F}{{1 - {F^2}}}} \right)^{\mu \rho }}{\left( {{\partial _\mu }F} \right)_{\rho \sigma }}{\left( {\frac{F}{{1 - {F^2}}}} \right)^{\sigma \nu }} = {\left( {{F^2} - 1} \right)^{\sigma \nu }}{\left( {\frac{1}{{1 - {F^2}}}} \right)^{\mu \rho }}{\left( {{\partial _\mu }F} \right)_{\rho \sigma }} + {\left( {\frac{1}{{1 - {F^2}}}} \right)^{\mu \rho }}{\left( {{\partial _\mu }F} \right)_{\rho \sigma }} = {\left( {\frac{1}{{1 - {F^2}}}} \right)^{\mu \rho }}{\left( {{\partial _\mu }F} \right)_{\rho \sigma }}{\left( {\frac{1}{{1 - {F^2}}}} \right)^{\sigma \nu }} - {\partial _\mu }F_\alpha ^\mu {\left( {\frac{1}{{1 - {F^2}}}} \right)^{\alpha \nu }} \\ \\ \end{gathered}$$

Comments (I'll sort out the formatting later.)

@user26143 : F does not commute with ∂μF, and one may demonstrate your formula only from the formula (53) of the matrix cookbook. – Trimok 16 hours ago

@Trimok, Excuse me, (53) in the matrix cookbook implies anticommuting relation, but not commuting. If I write [Fαβ,∂μFγδ], by Fαβ=∂αAβ−∂βAα and computing the eight terms after expanding the commutator, it seems to be zero. Or am I mistaken? – user26143 14 hours ago

@user26143 : For a general classical matrix F(x), ∂μF(x) does not commute with F(x). Here I have supposed that F(x)μν=Fμν(x) represents classical fields. – Trimok 14 hours ago

@user26143 : In the non-rigourous part of the answer, it is supposed that n and F commute, but it is not correct. – Trimok 14 hours ago

@Trimok, OK. I will edit the post ASAP, that in problem 5a Fμν=∂μAν−∂νAμ. – user26143 14 hours ago

@user26143 : I have a counterexample : By taking Ay=x2/2,Az=y, I found that the matrices F and ∂xF do not commute. – Trimok 14 hours ago

@user26143 : I must leave now. If you agree with me, I will provide an answer. If I am mistaken, you will tell me. – Trimok 14 hours ago

@Trimok Thank you very much! What is Ax in your example? I have to leave too, sorry – user26143 14 hours ago

@user26143 : A0=0,Ax=0,Ay=x2/2,Az=y , F=⎛⎝⎜⎜000000−x00x0−10010⎞⎠⎟⎟, ∂xF=⎛⎝⎜⎜000000−1001000000⎞⎠⎟⎟ – Trimok 11 hours ago

@Trimok, I agree with your counterexample. I cannot use [Fαβ,∂μFγδ]=0 to infer [F,∂μF]=0. I am grateful to see your approach, at least any hint – user26143 10 hours ago

@user26143 : I gave an answer – Trimok 10 hours ago

@Trimok, Thanks a lot!! I didn't think about the inverse of A/B – user26143 10 hours ago 1

@user26143 : you must be careful, there is an ambiguity with AB, because it could be B−1A or AB−1, and BA,could be A−1B or BA−1. However, in your case, it works in all cases, that is the inverse of F1−F2 is 1−F2F, whatever convention you are choosing. – Trimok 10 hours ago 1

@user26143 : You can check this by writing F1−F2=F(1+F2+F4+...+F2n+...)=(1+F2+F4+...+F2n+...)F and 1−F2F=F−1(1−F2)=(1−F2)F−1 – Trimok 10 hours ago upvote flag Don't forget to edit the relevant information from this comment thread into the answer, and then you can flag the comments for deletion. – David Z♦ 4 hours ago

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How did you get $F^2$ term in your first line in the "Rigorous proof"? I tried to write the term explicitly from the second term in Eq. (3), $$\left( \frac{F}{1-F^2} \right)^{\mu\rho} \left( \frac{F}{1-F^2} \right)^{\sigma \nu} = F^{\mu}_{\lambda} \left( \frac{1}{1-F^2} \right)^{\lambda \rho} F^{\sigma}_{\eta} \left( \frac{1}{1-F^2} \right)^{\eta \nu} - {\delta}^{\sigma}_{\eta} \left( \frac{1}{1-F^2} \right)^{\lambda \rho} {\delta}^{\sigma}_{\eta} \left( \frac{1}{1-F^2} \right)^{\eta \nu}$$ The factors $F^{\mu}_{\lambda}$ and $F^{\sigma}_{\eta}$ do not form $F^2$ by contracting indices? – user26143 Sep 1 '13 at 10:41
@user26143: There's neither a delta, nor an eta in my answer. Read it again. And the rigorous proof is equivalent to the non-rigorous one, I just don't let the indices take care of themselves. – centralcharge Sep 1 '13 at 10:44
I mean, how did you get $(F^2-1)^{\sigma \nu}$? I omitted the superscript ${\sigma \nu}$ previously. I am not saying if you have a delta/eta in your post, I tried to get your equation from my Eq.(3), but I cannot reach it. That is my step. – user26143 Sep 1 '13 at 10:45
@user26143 : you must be careful, there is an ambiguity with $\frac{A}{B}$, because it could be $B^{-1}A$ or $AB^{-1}$, and $\frac{B}{A}$,could be $A^{-1}B$ or $BA^{-1}$. However, in your case, it works in all cases, that is the inverse of $\frac{F}{1-F^2}$ is $\frac{1-F^2}{F}$, whatever convention you are choosing. – Trimok Sep 1 '13 at 19:04
@user26143 : You can check this by writing $\frac{F}{1-F^2} = F(1+F^2+F^4+...+F^{2n}+...) = (1+F^2+F^4+...+F^{2n}+...)F$ and $\frac{1-F^2}{F} = F^{-1}(1-F^2) = (1-F^2)F^{-1}$ – Trimok Sep 1 '13 at 19:08