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Talk:Dedekind eta function

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Latest comment: 3 years ago by SteJaes in topic Fractals?

Fractals?

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Why is this article in the "fractals" category? —Preceding unsigned comment added by 94.170.104.35 (talk) 22:58, 15 July 2009 (UTC)Reply

If You have ever seen the function as a correct 3D complex surface plot, You would not ask. So I want to impress and show a plot by Wolfram Mathematica:
Dedekind eta function over the complex numbers (high resolution)
Dedekind eta function over the complex numbers (high resolution)
SteJaes (talk) 14:19, 30 April 2023 (UTC)Reply

Typo/question

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The general modular transformation gives the eta function an argument z on the r.h.s., \tau would be more readable. (anon poster Dec 2005)

Yes that's a typo, now fixed linas 23:30, 16 December 2005 (UTC)Reply

\epsilon(a,b,c,d) seems to be independent of b, is this right? (anon poster Dec 2005)

Yes, but that's the standard notation. Besides, ad-bc=1 so the four args aren't independent anyway. linas 23:30, 16 December 2005 (UTC)Reply

The text mentions that eta is a modular function of weight 1/2. It'd be useful to include its level, and possibly character as well.

I fixed the definition

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I think the definition of eta shown here was wrong and instead of q^(1/24) it should be exp(i pi tau / 12) as I've made it. If the latter expression was implied in the first place, then that should be mentioned since the two expressions are ordinarily different.

The former expression, q^(1/24) led to some problems: 1. Take tau=1/2 + i * sqrt(3)/2. Then -1/tau = -1/2 + i * sqrt(3/2) = tau - 1. Note that q corresponding to that tau is -exp(-pi*sqrt(3)), while the q corresponding to -1/tau = tau - 1 is the exact same value. Thus the Dedekind eta function has the same value at both tau and -1/tau. But then the functional equation says that eta(-1/tau) = sqrt(-i*tau) * eta(tau), i.e. 1 = sqrt(-i*tau). This is incorrect. 1 is not the square root of (-i/2 + sqrt(3)/2).

2. Another problem is when tau = -1/2 + 0.01i, then eta(tau+1) = eta(tau) and the exp(i pi/12) factor couldn't be correct. Doubledork (talk) 23:02, 13 January 2012 (UTC)Reply

Strictly speaking, if the Dedekind eta function is seen as the 24th root of , the numerator being the modular discriminant, then the old definition was still correct. One can validly affix the appropriate 24th root of unity to so it can distinguish from . However, for the sake of clarity, it does seem reasonable to use instead of q^{1/24}, and I have made the necessary changes.Titus III (talk) 16:01, 5 March 2012 (UTC)Reply