# Why do those two terms add here?

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1. Dec 2, 2015

### Emilie.Jung

When I was studying complex manifolds in a Freedman's SUGRA book, I ran across this.

In a complex manifold, the metric is

$$ds^2=g_{ab}dz^adz^b\hspace{3cm}(1)$$

$$ds^2=2g_{\alpha\bar{\beta}}dz^{\alpha}d\bar{z}^{\bar{\beta}}+g_{\alpha\beta}dz^{\alpha\beta}dz^{\alpha}dz^{\beta}+g_{\bar{\alpha}\bar{\beta}}d\bar{z}^{\bar{\alpha}}d\bar{z}^{\bar{\beta}}\hspace{3cm}(2)$$

It occurred in his introductory chapter which must have been there to build the ground for coming chapter which has to do with special geometry.

My question here is about the first term in (2), why does it have a "2" multiplied by $g_{\alpha\bar{\beta}}dz^{\alpha}d\bar{z}^{\bar{\beta}}$? Is it because some two terms add giving this term. If so, what are those two terms and why do they add?

2. Dec 3, 2015

### mathman

The dz terms in the product commute. Breaking each into real and imaginary parts and forming the product gives you 2 terms of a real part times an imaginary part.

Note that the second term of your second expression has an obvious error - an extra dz term.

3. Dec 12, 2015

### Emilie.Jung

Oh sorry @mathman I didn't pay attention to the alert for your input on this thread! Thank you for pointing out the error, indeed it must have been $g_{\alpha\beta}dz^{\alpha}dz^{\beta}$ instead.

Now, regarding your reply, did you mean that the first term is the addition of $g_{\alpha\bar{\beta}}dz^{\alpha}d\bar{z}^{\bar{\beta}}+g_{\bar{\alpha}\beta}d\bar{z}^{\bar\alpha}dz^{\beta}$? If so, you then meant that since $g_{\alpha\bar{\beta}}=g_{\bar{\alpha}\beta}$ and since $dz^{\alpha}d\bar{z}^{\bar{\beta}}=d\bar{z}^{\bar\alpha}dz^{\beta}$ then this holds?

4. Dec 12, 2015

### nrqed

Yes, your explanation is indeed the correct one.

5. Dec 13, 2015

### Emilie.Jung

@nrqed Thank you for pointing it out as correct. I just am wondering why is that those can be added together?

Please note that as I was looking more into this on google https://books.google.com.lb/books?id=sfUICAAAQBAJ&pg=PA44&lpg=PA44&dq=we+now+define+a+hermitian+manifold+as+a+complex+manifold+where+there+is+preferred+class+of+coordinate&source=bl&ots=lS_eMIl13j&sig=arjL5QBrRexp5382VbyltH6_C5U&hl=ar&sa=X&ved=0ahUKEwjdmIHI0dfJAhVBUhoKHQJ_AbwQ6AEIJzAB#v=onepage&q=we now define a hermitian manifold as a complex manifold where there is preferred class of coordinate&f=false, I found something similar explaining this in a more detailed way:

"We now define a Hermitian manifold as a complex manifold where there is a preferred class of coordinate systems in which unmixed components of metric tensor vanish ($g_{\alpha\beta}=g_{\bar{\alpha}\bar{\beta}}=0$). The line element then takes the form $ds^2=2g_{\alpha\bar{\beta}}dz^{\alpha}d\bar{z}^{\beta}.$Coordinate systems for which this form holds are said to be adapted to the Hermitian structure. Hermitean metrics are definitely a restriction. Readers can show this by formlating the equations for a coordinate transform from a general line element $ds^2=g_{AB}dz^Adz^B$ to a Hermitan one."

So, it looks like imposed Hermiticity, so he moved from the general form $ds^2=g_{AB}dz^Adz^B$ to the final form $ds^2=2g_{\alpha\bar{\beta}}dz^{\alpha}d\bar{z}^{\beta}$. Is that right? Was hermiticity the missing block? If so, this contradicts with what we agreed on last, right?

6. Dec 13, 2015

### nrqed

I don't see any contradiction with what we agreed on last night. We have $ds^2 = g_{\alpha \beta } dz_\alpha dz_\beta + g_{\bar{\alpha}\bar{\beta}} dz_{\bar{\alpha}} dz_{\bar{\beta}} + g_{\bar{\alpha} \beta} dz_{\bar{\alpha}} dz_{\beta} + g_{\alpha \bar{\beta}} dz_{\alpha} dz_{\bar{\beta}}$. By hermiticity the first two terms are zero and from your argument last night , the last two can be combined (keeping in mind that the indices are summed over, we see that the last two terms are equal)

7. Dec 13, 2015

### Emilie.Jung

When last night I said
I didn't know that because of Hermiticity that's the case. I was just following up @mathman 's answer, but without knowing why he considered that to be the case. He said

So why
Maybe I didn't precisely understand what you meant by

8. Dec 13, 2015

### nrqed

I am sorry, I had misread your post here. We do not have to assume $g_{\alpha\bar{\beta}}=g_{\bar{\alpha}\beta}$, all we need is to use is symmetry of the metric,
$g_{\alpha\bar{\beta}}=g_{\bar{\beta}\alpha}$.

What I meant with my comment about summation is this. Let's say we fix $\alpha, \beta = 1,2$. If we consider

$$g_{\alpha\bar{\beta}}dz^{\alpha}d\bar{z}^{\bar{\beta}}+g_{\bar{\alpha}\beta}d\bar{z}^{\bar\alpha}dz^{\beta} = g_{1,\bar{2}} dz^1 dz^{\bar{2}} + g_{\bar{1},2}dz^{\bar{1}} dz^2$$

then we cannot simplify this in any way, the two terms cannot be combined. In other words, we may not use here $g_{1,\bar{2}} dz^1 dz^{\bar{2}} =g_{\bar{1},2}dz^{\bar{1}} dz^2$.

However, if we sum over all values of the indices, then we *can* write that the sum $g_{\alpha\bar{\beta}}dz^{\alpha}d\bar{z}^{\bar{\beta}}$ is indeed equal to $g_{\bar{\alpha}\beta}d\bar{z}^{\bar\alpha}dz^{\beta}$, as we can check by writing down all the terms and using the symmetry of the metric.

9. Dec 13, 2015

### Emilie.Jung

You meant by
that if I write

$$ds^2=g_{1\bar{1}}dz^1d\bar{z}^{\bar{1}}+g_{\bar{1}1}d\bar{z}^{\bar{1}}dz^1+ g_{2\bar{2}}dz^2d\bar{z}^{\bar{2}}+g_{\bar{2}2}d\bar{z}^{\bar{2}}dz^2+ g_{1\bar{2}}dz^1d\bar{z}^{\bar{2}}+g_{\bar{1}2}d\bar{z}^{\bar{1}}dz^2+ g_{2\bar{1}}dz^2d\bar{z}^{\bar{1}}+g_{\bar{2}1}d\bar{z}^{\bar{2}}dz^1$$
then we say because of hermiticity the first 4 terms vanish and then we combine (5th term with the 8th term) and (the 6th term with the 7th term) to get

$$ds^2= 2g_{2\bar{1}}dz^2d\bar{z}^{\bar{1}}+2g_{1\bar{2}}dz^1d\bar{z}^{\bar{2}}?$$ but I'm left with 2 terms now, so what should be done to reach the form we're seeking which is only
$$ds^2=2g_{\alpha\bar{\beta}}dz^{\alpha}d\bar{z}^{\bar{\beta}}?$$

10. Dec 13, 2015

### nrqed

Not quite. You have already used hermiticity because you did not write any of the terms from $g_{\bar{\alpha}\bar{\beta}}dz^{\bar{\alpha}}d\bar{z}^{\bar{\beta}}+ g_{\alpha\beta}dz^{\alpha} dz^{\beta}$

I am not sure I follow the rest of your post. Let me go back to what you wrote above.

Note that

The first term and the second term are identical

The third and fourth terms are identical

the 5th and 8th terms are equal, as you said

the 6th and 7th term are equal, as you said

So they all add up and give the same thing as if we had written
$2g_{\alpha\bar{\beta}}dz^{\alpha}d\bar{z}^{\bar{\beta}}$

11. Dec 13, 2015

### Emilie.Jung

@nrqed Right, I noticed where I had gone wrong! Thank you a lot for your assistance and great help!

12. Dec 13, 2015

### nrqed

Good! An you are welcome.

13. Dec 13, 2015

### samuelphysics

@nrqed But why do the dz^i terms commute?

14. Dec 13, 2015

### nrqed

They are just ordinary differentials, this is not a wedge product of one-forms.

15. Dec 13, 2015

### samuelphysics

The past post got a little messy, what I meant to ask you was when you said
So why is it that hermiticity makes those two terms vanish? Excuse me for double posting

16. Dec 13, 2015

### Emilie.Jung

@samuelphysics I guess this is the definition of a hermitian metric, that the unmixed components of it vanish, I thus think it is a property. I'm not sure about its proof, though. I don't want to be giving you unsure or wrong answers, though.

@nrqed is this right?

17. Dec 13, 2015

### nrqed

Samuel physics , As EmilieJung said, this is not true of a general metric, here you must see it as the question being about a special class of metric with that extra condition. There is nothing more to it, as far as this thread is concerned. Why they are interesting is a good question but an entirely different one :-)

18. Dec 13, 2015

### samuelphysics

ummm, thanks alot @Emilie.Jung and @nrqed ! So, you mean this is not a property of a Hermitian metric but is an additional condition on that Hermitian metric? Because EmilieJung said it was a property. If EmilieJung is right then doesn't this property have a proof? That was mainly my question.

19. Dec 16, 2015

### Emilie.Jung

@nrqed I want to ask you about this because @samuelphysics's last question got me a little confused. We say that a Hermitian manifold is one
in which unmixed components of metric tensor vanish ($g_{\alpha\beta}=g_{\bar{\alpha}\bar{\beta}}=0$). Is that correct? So every Hermitian metric there is has this special property, no? I just want to double check from you.