# constants in Gamma factors in functional equation for zeta functions.

Usually the Riemann zeta function $\zeta(s)$ gets multiplied by a "gamma factor" to give a function $\xi(s)$ satisfying a functional equation $\xi(s)=\xi(1-s)$. If I changed this gamma factor by a non-zero constant complex number, the functional equation would still hold.

More generally, for zeta functions of number fields, one gets a gamma factor for each real place and a gamma factor for each complex place. In a parallel universe, we could have defined the gamma factor for each real place to be (say) $\pi^{1/2}$ times what we usually use, and the gamma factor at the complex places to be (say) $1/2$ of what we usually use (and I think I was taught to use $2.(2\pi)^s$ so we could just knock off that first 2) and all functional equations would still be exactly the same (because the extra factors would be the same on both sides).

More generally, for Dirichlet $L$-functions and Hecke $L$-functions: I now need a gamma factor for the sign function on the non-zero reals, and again we could use a different choice.

Now some general yoga of gamma factors tells us that really there are only 3 choices to be made (because Hodge structures basically always decompose into the types covered above)---and I've mentioned them all already.

Upshot: did we, at some point, make 3 arbitrary choices of constants, and the entire theory of $L$-functions of motives wouldn't care what choices we made, so we could have made other choices? Note for example that conjectures on special values of $L$-functions don't take the gamma factors into account (well, the ones I know don't; they predict values of the incopleted $L$-function without the gamma factors). Note also that when defining these things for Hecke characters a la Tate, again pretty arbitrary choices are made at the infinite places---there is no one canonical function that is its own Fourier transform, because we can change things by constants again.

Am I totally wrong here or are there really 3 arbitrary choices that we have made, and we could have made others?

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The gamma factor at a real place already considers the sign function (unless I'm misunderstanding what you mean), on the other hand, you've probably also made a choice of square root of -1. – Rob Harron Feb 4 '10 at 21:28
@Rob H: What I was trying to say about real places could be summarised like this: the zeta function has Gamma(s/2) in its Euler factor, and for an odd Dirichlet character it has Gamma((s+1)/2) in. And I'd like to think that I'd not made a choice of square root of -1...not yet at least. My Hecke characters are valued in C though, rather than R-bar (so I've made a choice in my coefficient field). – Kevin Buzzard Feb 4 '10 at 22:47
Hi Kevin. This is probably orthogonal to your question, but according to my understanding of Deninger's point of view, Gamma factors have a preferred normalization---the one that makes his regularized products work. See "On the Gamma-factors attached to motives" (Inv. Math. 1991). From what I remember, he asserts elsewhere that this normalization is supported by the point of view in Tate's thesis. (NB This is all based on a cursory reading.) – JBorger Feb 5 '10 at 2:51
I was going to ask if Deninger's work suggested a particular normalization, but Jim seems to have answered that! Do I remember correctly that in Jacquet--Langlands they just choose the standard Gamma factor by convention (after having shown that the local factor has to lie in the one-dimensional space that it spans, by some kind of Whittaker model computation)? – Emerton Feb 5 '10 at 3:48
@James: in the Jacquet-Langlands book they simply decree that at a non-arch place an Euler factor must be of the form P(q^{-s})^{-1} with P a polynomial with constant term 1. They give an "intrinsic" characterisation of the factor that gives it only up to a non-zero constant, and then they normalise it with this "constant term 1" trick. At the infinite places they basically give the same "intrinsic" characterisation, that only works up to a constant, and then they just decree what the answer is and note that this works fine. So there's a definite difference. – Kevin Buzzard Feb 9 '10 at 17:09