Today I Learned

March 12th

Today I learned the details of the proof of the local $\zeta$ functional equation. As fair warning, this is going to be quite involved. We would like to begin by defining our local $L$-factors, but for this, we have to talk about classifiaction of characters. A few days ago we showed the following.

We outline this. The idea is to write $\chi=\frac{\chi}{|\chi|}\cdot|\chi|.$ The first character is unitary (it outputs into $S^1$ for free). The second character $|\chi|$ is unramified (showing this is the hard part), so it has the form $x\mapsto|x|^s$ for some $s\in\CC.$ $\blacksquare$

We can quickly use this result to classify our multiplicative characters of $\RR$ and $\CC.$

Recall that a character $\chi$ is unramified if and only if it vanishes on the norm-$1$ elements of $\RR^\times,$ which here are only $\pm1.$ Because $\chi(1)=1$ surely, and $\chi(-1)^2=\chi(1)=1,$ we see $\chi(-1)=\pm1.$ This gives two cases.

• If $\chi(-1)=1,$ then $\chi$ is trivial on the norm-$1$ elements already, so $\chi$ is unramified, so we may take $a=0.$

• If $\chi(-1)=-1,$ then $x\mapsto x\chi(x)$ is a character trivial on the norm-$1$ elements, so we write $x\chi(x)=|x|^s$ for some $s\in\CC,$ and here $a=1.$

Having dealt with all cases, we are done here. $\blacksquare$

Before talking about $\CC,$ we mention that we will write $\lVert z\rVert=|z|^s$ to be the natural (but unusual) square of the normal absolute value. (The reason $\lVert\bullet\rVert$ is natural is because it is the norm induced by scaling space, but this is unimportant.)

As above, we focus on $\chi$'s behavior over norm-$1$ elements. Fix $\eta:=\chi|_{S^1}$ a multiplicative character on $S^1.$ Because $S^1$ is compact, $\eta$ is unitary automatically, so $\eta:S^1\to S^1.$

Now, $(S^1,\times)\cong(\RR/\ZZ,+)$ by sending $e^{2\pi i\theta}\mapsto\theta.$ (We are not going to be rigorous about this.) Thus, $\eta$ through this isomorphism induces a continuous homomorphism $\varphi:\RR/\ZZ\to S^1.$ By extending $\varphi$ periodically/pulling $\varphi$ back from $\RR/\ZZ,$ we get a continuous map $\varphi:\RR\to S^1.$ By our classification of characters on local fields, then, we get some $\alpha\in\RR$ for which\[\varphi(x):=e^{2\pi i\alpha x}.\]Because $\varphi$ needs to act trivially on $\ZZ$ to factor through $\RR/\ZZ,$ we know $\varphi(1)=e^{2\pi i\alpha}=1,$ so it follows $\alpha$ is an integer. Thus, $\varphi(x)=e^{2\pi inx}$ for some integer $n.$ Pulling back to $\eta,$ we know that\[\eta\left(e^{2\pi i\theta}\right)=\varphi(\theta)=e^{2\pi in\theta},\]so $\eta:z\mapsto z^n$ for the same integer $n.$

Having studied $\chi|_{S^1},$ we now expand back out to $\chi.$ Extend $\eta$ from $S^1$ to all of $\CC^\times$ by defining $\eta(z)=z^n$ for all $z\in\CC^\times.$ Then $\chi/\eta$ is well-defined, and furthermore,\[(\chi/\eta)\left(e^{i\theta}\right)=\frac{\chi\left(e^{i\theta}\right)}{\chi|_{S^1}\left(e^{i\theta}\right)}=1,\]so $\chi/\eta$ is trivial on $S^1$ and therefore unramified. So we can fix $\chi(z)=z^n\lVert z\rVert^s$ for some $s\in\CC.$

To convert this to the desired form, note that we just have $-a=n$ if $n$ is a nonpositive integer. Otherwise, $n$ is nonnegative, and we note that $z\overline z=\lVert z\rVert,$ so\[z^n\lVert z\rVert^s=\overline z^{-n}\cdot\overline z^nz^n\lVert z\rVert^s=\overline z^{-n}\lVert z\rVert^s,\]so we may take $b=n$ to finish. Having covered all cases of $n,$ we are done here. $\blacksquare$

We take a second to realize that the above classifications are really giving the components of the space of characters. Indeed, a decomposition $\chi(x)=\eta(x)|x|^s$ means that we can parameterize a component of characters with the same $\eta$ by the variable $s\in\CC,$ so we can talk about being holomorphic with respect to $s$ and similar. Each of the above classes give a different $\eta.$

• With $F=\RR,$ we note that $a=0$ has $\eta=1,$ and $a=1$ has $\eta(x)=x^{-1}|x|=\op{sgn},$ which are indeed distinct.

• With $F=\CC,$ we note that $a=b=0$ has $\eta=1,$ $a \gt 0$ has $\eta(z)=z^{-a}\lVert z\rVert^{a/2},$ and $b \gt 0$ has $\eta(z)=\overline z^{-b}\lVert z\rVert^{b/2}.$

We mention this now because we will refer to the above classifications later on even though it is arguably more natural to refer to the components by their $\eta.$

We are now ready to define our local $L$-factors. They are as follows.

These more or less correspond to the various parts of the completed Riemann $\xi_\QQ$ function, where we had a real $\Gamma_\RR$ $L$-factor from the archimedean completion and then a $\frac1{1-p^{-s}}$ for each $p,$ which corresponds to $L$-factors from the nonarchimedean completions.

Now, our statement for the local $\zeta$ functional equation is going to look like\[\frac{\zeta(\hat f,\chi^\vee)}{L(\chi^\vee)}=\varepsilon(\chi,\psi,dx)\frac{\zeta(f,\chi)}{L(\chi)}\]for any Shwartz-Bruhat $f,$ a multiplicative character $\chi,$ and some function $\varepsilon$ which depends on the character $\chi,$ our choice of standard character $\psi$ (for Fourier transforms), and choice of Haar measure $dx$ on our local field. Notably, $\varepsilon$ does not depend on our choice of multiplicative measure $d^\times x=c\cdot dx/|x|$ because this $c$ will cancel as it appears in both $\zeta$ factors.

Last time we showed that, after fixing a character $\chi,$ for Shwartz-Bruhat functions $f$ and $g,$ we have\[\zeta(f,\chi)\zeta(\hat g,\chi^\vee)=\zeta(\hat f,\chi^\vee)\zeta(g,\chi)\]by expanding everything out into a triple integral and creating symmetry. Thus, if we can exhibit a single $f$ for each $\chi$ for which the functional equation\[\frac{\zeta(\hat f,\chi^\vee)}{L(\chi^\vee)}=\varepsilon_f(\chi,\psi,dx)\frac{\zeta(f,\chi)}{L(\chi)},\]then we would like to rearrange this to\[\frac{\zeta(\hat g,\chi^\vee)}{L(\chi^\vee)}=\frac{\zeta(\hat g,\chi^\vee)}{\zeta(\hat f,\chi^\vee)}\cdot\frac{\zeta(\hat f,\chi^\vee)}{L(\chi^\vee)}=\frac{\zeta(g,\chi)}{\zeta(f,\chi)}\cdot\varepsilon\frac{\zeta(f,\chi)}{L(\chi)}=\varepsilon\frac{\zeta(g,\chi)}{L(\chi)}.\]This manipulation works as long as $\zeta(f,\chi)$ and $\zeta(\hat f,\chi)$ are nonzero, and because we will be primarily interested in the region where $\chi(x)=\eta(x)|x|^s$ with $\op{Re}(s)\in(0,1),$ we need these to be nonzero there.

The point now is that it suffices to show the functional equation for a single function $f$ for each character $\chi.$ This is still a bit of a daunting task, but it is made manageable because $\chi$ can be decomposed into $\chi(x)=\eta(x)|x|^s,$ and we will do a long casework on $\eta.$

Before continuing, we are going to do a lot of computations, so we remark that we are allowed to fix $\psi$ and $dx$ however we please, and though this will change $\varepsilon,$ it only changes things by a constant, maintaining the underlying result. Indeed, if we choose a different $\psi_a:x\mapsto\psi(ax)$ or different $c\cdot dx,$ then our Fourier transform changes to\[\hat f_{a,c}(y)=\int_Ff(x)\psi_a(xy)\,cdx=c\int_Ff(x)\psi(x(ay))\,dx=c\hat f(ay),\]so\[\zeta(\hat f_{a,c},\chi^\vee)=\int_{F^\times}\hat f_{a,c}(x)\chi^\vee(x)\,d^\times x=c\int_{F^\times}\hat f(ax)\chi^\vee(x)\,dx,\]which is\[\zeta(\hat f_{a,c},\chi^\vee)=\frac c{\chi^\vee(a)}\int_{F^\times}\hat f(ax)\chi^\vee(x)\,d^\times x=\frac c{\chi^\vee(a)}\zeta(\hat f,\chi^\vee)\]by taking $x\mapsto x/a.$ The point is that the functional equation\[\frac{\zeta(\hat f,\chi^\vee)}{L(\chi^\vee)}=\varepsilon(\chi,\psi,dx)\frac{\zeta(f,\chi)}{L(\chi)}\]will still hold with $\varepsilon(\chi,\psi_a,cdx)=\frac c{\chi^\vee(a)}\varepsilon(\chi,\psi,dx).$ Indeed, $\zeta(\hat f,\chi^\vee)$ is the only term affected by altering $\psi$ and $dx$ (changing $dx$ multiplies both $d^\times x$ factors in the $\zeta$s by the same factor), and our $\varepsilon$ has only needed to be multiplied by a nonzero factor to account for this. So we are allowed to choose our $\psi$ and $dx$ conveniently.

With that said, we begin with $F=\RR.$

We make $dx$ the Lebesgue measure and $\psi(x)=e^{-2\pi ix}$ so that $g_\RR(x):=e^{-\pi x^2}$ satisfies $\widehat{g_\RR}=g_\RR,$ which we won't prove here; here is a reference. (The point is that the measure is chosen to make this function self-dual.) Additionally, we use $d^\times x=dx/|x|$ for convenience. Now, fix $\chi:x\mapsto x^{-a}|x|^s$ as described in our classification.

We want to make our choice of $f$ as close to self-dual as possible, in order to minimize headaches, but we also want to make the integral $\zeta(f,\chi)$ relatively unpainful. The solution is to let $f(x)=x^ae^{-\pi x^2}$ in order to offset the $x^{-a}$ introduced by $\chi.$ This gives\[\zeta(f,\chi)=\int_{\RR^\times}f(x)\chi(x)\,d^\times x=\int_{\RR^\times}x^ae^{-\pi x^2}\cdot x^{-a}|x|^s\,d^\times x.\]Here, the $x^a$ and $x^{-a}$ do indeed cancel. We would like to make this look like $L(\chi)=\Gamma_\RR(s)=\pi^{-s/2}\Gamma(s/2),$ so we will need to make $\Gamma(s)=\int_0^\infty e^{-t}t^s\,d^\times t$ sometime soon. So we message the above into\[\zeta(f,\chi)=2\int_0^\infty e^{-\pi x^2}x^s\,d^\times x\]and then substitute $t:=\pi x^2$ so that $dt/t=2\pi x/\left(\pi x^2\right)=2d^\times x,$ giving\[\zeta(f,\chi)=\int_0^\infty e^{-t}\left(\frac t\pi\right)^{s/2}\,d^\times t,\]which does indeed equal $\pi^{-s/2}\Gamma(s/2)=L(\chi),$ giving $\zeta(f,\chi)/L(\chi)=1.$ We also quickly note that $\zeta(f,\chi)$ is nonzero because $\Gamma(s)$ has no zeroes.

It remains to deal with $\hat f,$ for which some care is required. We do casework on $a.$

• If $a=0,$ then $f:=g_\RR$ is self-dual. Thus $\zeta(\hat f,\chi^\vee)=\zeta(f,\chi^\vee)=L(\chi^\vee)$ from the work above. So $\zeta(\hat f,\chi^\vee)$ is still nonzero, and we can simply use $\varepsilon=1$ here.

• If $a=1,$ then $f(x)=xe^{-\pi x^2}.$ To compute this Fourier transform, we start by noting we already have \[e^{-\pi y^2}=g_\RR(y)=\widehat{g_\RR}(y)=\int_\RR e^{-\pi x^2}e^{-2\pi ixy}\,dx.\] The key trick is to take $\del/\del y$ by differentiating under the integral, giving \[-2\pi ye^{-\pi y^2}=-2\pi i\int_\RR xe^{-\pi x^2}e^{-2\pi ixy}\,dx.\] The left-hand side is $-2\pi f(y),$ and the right-hand side is $-2\pi i\hat f(y),$ so we see $\hat f(y)=-if(y).$ Because $\chi^\vee(x)=\chi(x)^{-1}|x|=x|x|^{-s+1}=x^{-1}|x|^{3-s}$ has the same form as $\chi$ (is in the same component), our earlier argument still gives \[\zeta(\hat f,\chi^\vee)=\zeta(-if,\chi^\vee)=-i\zeta(f,\chi^\vee)=-iL(\chi^\vee),\] where the second equality holds because constants can move through the $\zeta$ integral. In particular, it follows that $\zeta(\hat f,\chi^\vee)/L(\chi^\vee)=-i,$ so $\varepsilon=-i$ will work. And $\zeta(\hat f,\chi^\vee)$ is still nonzero because $\zeta(f,\chi^\vee)$ is.

Having dealt with all cases, having exhibited our $\varepsilon$ and shown that the $\zeta$ terms are nonzero in the area of interest, we are done. $\blacksquare$

Next we do $F=\CC.$ The statement we prove is pretty much identical.

We make $dz$ twice (!) the Lebesgue measure and $\psi(z)=e^{-2\pi i(z+\overline z)}$ so that $g_\CC(x):=e^{-2\pi z\overline z}$ is self-dual; again, we won't show this here. We also take $d^\times z=dz/\lVert z\rVert$ for convenience.

Fix $\chi(z)=z^{-a}\overline z^{-b}\lVert z\rVert^s$ as described above. Here, we have one of $a=0$ or $b=0,$ but because $z\mapsto\overline z$ is an automorphism, we can just take $b=0$ without loss of generality because writing $z\mapsto\overline z$ in the below argument won't change anything.

Again, we want our $f$ to be almost self-dual while making the $\zeta(f,\chi)$ not too painful, so we do the same trick we had in the previous proposition by taking $f(z)=z^ag_\CC(z).$ Now the computation begins. We have\[\zeta(f,\chi)=\int_{\CC^\times}f(z)\chi(z)\,d^\times z=\int_{\CC^\times}z^ae^{-2\pi z\overline z}\cdot z^{-a}\lVert z\rVert^s\,d^\times z.\]Here the $z^a$ and $z^{-a}$ will cancel. Moving this into an integral over $\RR,$ we let $z=x+yi$ so that $dz=2dxdy.$ This gives\[\zeta(f,\chi)=\int_{\RR^2}e^{-2\pi\left(x^2+y^2\right)}\left(x^2+y^2\right)^s\cdot\frac{2dxdy}{x^2+y^2}.\]We quickly explicitly observe that $z\mapsto\overline z$ taking $x+yi\mapsto x-yi$ will not change this integral at all, so our without loss of generality above is justified. In fact, this is entirely a function $x^2+y^2,$ so we transfer this integral into polar, giving\[\zeta(f,\chi)=\int_0^{2\pi}\int_0^\infty e^{-2\pi r^2}r^{2s}\cdot\frac{2rdrd\theta}{r^2}.\]We can move the $d\theta$ integral outside to give a factor of $2\pi$ in the front. Like with $\RR,$ we would again like to make $\Gamma(s)=\int_0^\infty e^{-t}t^s\,d^\times t$ appear, so we let $t=2\pi r^2$ so that $d^\times t=dt/t=4\pi rdr/\left(2\pi r^2\right)=2dr/r.$ This gives\[\zeta(f,\chi)=2\pi\int_0^\infty e^{-t}\left(\frac t{2\pi}\right)^s\,d^\times t,\]which collapses into $2\pi(2\pi)^{-s}\Gamma(s)=\pi\Gamma_\CC(s)=\pi L(\chi).$ Again, this $\zeta(f,\chi)$ because $\Gamma$ has no zeroes.

Next we compute the Fourier transform of $f_a(z):=z^ae^{-2\pi z\overline z}.$ This is a bit involved and out-of-the-way, so we put it in its own lemma.

This is mostly a computational bash, but Tate has a trick to make this not too painful. We proceed by induction. For $a=0,$ this is simply asserting that $f_0=g_\CC$ is self-dual, which we continue to assert without proof. (It comes down to a full expansion into integrals in $\RR$ and then using self-duality of $g_\RR.$)

For the inductive step, we suppose the statement for $f_a$ and will compute the Fourier transform of $f_{a+1}.$ We begin by writing\[\widehat{f_a}(t)=\int_\CC f_a(z)e^{-2\pi(zt+\overline{zt})}\,dz=\int_\CC z^ae^{-2\pi z\overline z}\cdot e^{-2\pi(zt+\overline{zt})}\,dz.\]We don't understand integrals in $\CC$ very well, especially with our funny measure $dz,$ so we dissolve this into an integral over real variables by writing $z=x+yi$ and $t=u+vi.$ Then we note $zt+\overline{zt}=2\op{Re}(zt)=2(ux-vy),$ which gives\[\widehat{f_a}(u+vi)=\int_{\RR^2}(x+yi)^ae^{-2\pi\left(x^+y^2\right)}e^{-2\pi(ux-vy)}\,2dxdy.\]The idea that Tate has is to generalize the trick we used earlier of differentiating under the integral sign, but now we need to make $(x+yi)^a$ into $(x+yi)^{a+1}$ for the induction, and we also have two variables to deal with. To start, we write\begin{align*} \frac{\del\widehat{f_a}(u+vi)}{\del u} &= -2\pi\int_{\RR^2}x(x+yi)^ae^{-2\pi\left(x^+y^2\right)}e^{-2\pi(ux-vy)}\,2dxdy, \\ \frac{\del\widehat{f_a}(u+vi)}{\del v} &= +2\pi\int_{\RR^2}y(x+yi)^ae^{-2\pi\left(x^+y^2\right)}e^{-2\pi(ux-vy)}\,2dxdy.\end{align*}Thus, we see that\[\left(\frac\del{\del u}-i\frac\del{\del v}\right)\widehat{f_a}(u+vi)=-2\pi\widehat{f_{a+1}}(u+vi)\]because the integrals will accumulate an extra $x+yi$ term. However, we can use this statement to evaluate $\widehat{f_{a+1}}$ directly. Indeed, we see $\del\widehat{f_a}/\del u$ and $\del\widehat{f_a}/\del v$ are\begin{align*} \frac\del{\del u}(-i)^a(u-vi)^ae^{-2\pi\left(u^2+v^2\right)}= &+(-i)^aa(u-vi)^{a-1}e^{-2\pi\left(u^2+v^2\right)} \\ &+(-i)^a(u-vi)^a\cdot-2\pi ue^{-2\pi\left(u^2+v^2\right)}, \\ \frac\del{\del v}(-i)^a(u-vi)^ae^{-2\pi\left(u^2+v^2\right)}= &-i(-i)^aa(u-vi)^{a-1}e^{-2\pi\left(u^2+v^2\right)} \\ &+(-i)^a(u-vi)^a\cdot-2\pi ve^{-2\pi\left(u^2+v^2\right)}.\end{align*}by using the product rule. The first terms of both will cancel when computing $\frac\del{\del u}-i\frac\del{\del v}$ because $(-i)^2=-1.$ Then the second terms will combine into\[\left(\frac\del{\del u}-i\frac\del{\del v}\right)\widehat{f_a}(u+vi)=-2\pi(u-vi)\cdot(-i)^a(u-vi)^ae^{-2\pi\left(u^2+v^2\right)}.\]Cancelling the $-2\pi$ with its appearance in $-2\pi\widehat{f_{a+1}}(u+vi),$ we do indeed get\[\widehat{f_{a+1}}(u+vi)=(-i)^a(u-vi)^{a+1}e^{-2\pi\left(u^2+v^2\right)}.\]Setting $z=u+vi$ again, this is exactly we wanted, completing our inductive step. $\blacksquare$

The work we showed earlier gave us that $\zeta(f,\chi)=\pi L(\chi)$ for $f(z)=z^ae^{-2\pi z\overline z}$ and $\chi(z)=z^{-a}\lVert z\rVert^s,$ and we remarked that the same argument implies this for $f(z)=\overline z^be^{-2\pi z\overline z}$ and $\chi(z)=\overline z^{-b}\lVert z\rVert^s.$ Thus, $\hat f(z)=(-i)^a\overline z^ae^{-2\pi z\overline z}$ with $\chi^\vee=z^a\lVert z\rVert^{-s}\lVert z\rVert=\overline z^{-a}\lVert z\rVert^{a-s+1}$ match and give\[\zeta(\hat f,\chi^\vee)=(-i)^a\cdot\pi L(\chi),\]and again this is nonzero because $L(\chi)$ is made up of $\Gamma$ and itself nonzero. It follows that we are done with $\varepsilon=(-i)^a.$ $\blacksquare$

We now move on to nonarchimedean $F$; let $\mathcal O$ be its ring of integers and $\mf p$ its maximal ideal. While we're here, we also set $\varpi$ a uniformizer and $q:=[\mathcal O:\mf p]$ the size of the residue field, though we won't use them yet. In what follows, we will not use the standard measures but rather convenient ones: we use $dx$ which gives $\int_\mathcal Odx=1$ and $d^\times x=dx/|x|.$

We need to pick up a technical lemma, so we will do it now to take a bit of a break. We have the following definition.

To compare, the usual Gauss sum is done over $\FF_p$ and looks like\[g(\chi):=\sum_{x\in\FF_p^\times}\chi(x)\zeta_p^x,\]where $\chi:\FF_p^\times\to S^1$ is a multiplicative character. The above is the generalization from $\FF_p$ to the nonarchimedean case: here $\chi$ becomes $\omega$ and $x\mapsto\zeta_p^x$ is $\psi.$ As some pretentious commentary, we can technically make this definition work with the archimedean case, and indeed\[\Gamma(s)=\frac12\int_{\RR^\times}t^se^{-t}\,d^\times t\]matches the format pretty much exactly. So indeed, a lot of the work we've been doing is thinly veiled under studying the interaction (under integration) of additive and multiplicative characters.

Anyways, in the same way that we can talk about $|g(\chi)|^2$ in the finite field case, we can talk about $|g(\omega,\psi)|^2$ in the general nonarchimedean case.

The conductor is the largest ideal such that each character acts trivially. Namely, $\omega|_{1+\mf p^n}=1$ and $\psi|_{\mf p^m}=1,$ but no smaller $n$ or $m$ (i.e., no larger $1+\mf p^n$ or $\mf p^m$) will do.

We remark that the $|g(\omega,\psi)|^2$ in the statement is not using the squared absolute value but the usual one. Indeed, we have\[|g(\omega,\psi)|^2=g(\omega,\psi)\overline{g(\omega,\psi)}=\int_{\mathcal O^\times}\omega(x)\psi(x)\,d^\times x\cdot\overline{\int_{\mathcal O^\times}\omega(y)\psi(y)\,d^\times y}.\]Everything here absolutely converges (these are characters being integrated over a bounded subset of $F,$ so it converges), so we may freely write this as\[|g(\omega,\psi)|^2=\iint_{\left(\mathcal O^\times\right)^2}\omega(x)\overline{\omega}(y)\cdot\psi(x)\overline\psi(y)\,d^\times yd^\times x.\]Using the additivity/multiplicativity of our characters, the integrand is $\omega(x/y)\psi(x-y).$ Now, our measure is multiplicative, so it will be easiest to fix $t:=x/y$ with $d^\times t=d^\times x$ to simplify things. Additionally, $\mathcal O^\times$ is a multiplicative group, so this transformation does not change the domain, and we see\[|g(\omega,\psi)|^2=\iint_{\left(\mathcal O^\times\right)^2}\omega(t)\psi(yt-y)\,d^\times yd^\times x,\]which is\[|g(\omega,\psi)|^2=\int_{\mathcal O^\times}\omega(t)\left(\int_{\mathcal O^\times}\psi((t-1)y)\,d^\times y\right)d^\times t.\]Now that we have extricated one of our characters from the integral, it remains to investigate the inner integral.

Note that $y\mapsto\psi((t-1)y)$ is just another additive character, and to make life easier, we would like to make everything additive: note $d^\times y=dy/|y|=dy$ because $|y|=1$ currently, and then\[\int_{\mathcal O^\times}\psi((t-1)y)\,dy=\int_{\mathcal O}\psi((t-1)y)\,dy-\int_{\mf p}\psi((t-1)y)\,dy.\]Continuing with the observation that $y\mapsto\psi((t-1)y)$ is just another additive character, we note that it is trivial exactly on $\mf p^\bullet$ if and only if $(1-t)\mf p^\bullet\subseteq\mf p^m$ because $\mf p^m$ is the conductor of $\psi.$ This is equivalent to $t\in1+\mf p^{m-\bullet},$ so the integral collapses into\[1_{t\in1+\mf p^m}\int_{\mathcal O}dy-1_{t\in1+\mf p^{m-1}}\int_{\mf p}dy=1_{t\in1+\mf p^m}-1_{t\in1+\mf p^{m-1}}\frac1q.\]We note $\int_{\mf p}dy=1/q$ because $\mf p$ is one of $[\mathcal O:\mf p]=q$ cosets whose measure sum to $\int_{\mathcal O}dy=1.$

Returning to $|g(\omega,\psi)|^2,$ we see we only care about the $t$ with $t\in1+\mf p^m$ or $t\in1+\mf p^{m-1}.$ Namely, we have\[|g(\omega,\psi)|^2=\int_{1+\mf p^m}\omega(t)\,d^\times t-\frac1q\int_{1+\mf p^{m-1}}\omega(t)\,d^\times t.\]Only now do we do casework on $m$ and $n.$

• If $m \gt n,$ then $1+\mf p^n$ is a superset of both $1+\mf p^{m-1}$ and $1+\mf p^n,$ so $\omega$ is trivial in both integrals. However, we can then evaluate these integrals directly, for $d^\times t=dt/|t|=dt$ here, giving \[\int_{1+\mf p^\bullet}d^\times t=\int_{1+\mf p^\bullet}dt=\left[\mathcal O:\mf p^\bullet\right]^{-1}=q^{-\bullet}\] because $1+\mf p^\bullet$ is just one of $\left[\mathcal O:\mf p^\bullet\right]$ cosets. Thus, the expression for $|g(\omega,\psi)|^2$ turns into $q^{-m}-q^{-1}\cdot q^{-(m-1)}=0.$

• If $m=n,$ then $1+\mf p^n$ contains $1+\mf p^m$ but not $1+\mf p^{m-1},$ so the first integral is trivial and gives $q^{-m}$ while the second integral vanishes because $\omega$ is a nontrivial character. So we get $q^{-m}.$

• If $m \lt n,$ then $1+\mf p^n$ is contained in both $1+\mf p^m$ and $1+\mf p^{m-1},$ so $\omega$ is a nontrivial character in both integrals, implying that they both vanish to $0.$

Combining all cases, we see that $|g(\omega,\psi)|^2$ is $q^{-m}$ when $m=n$ and $0$ otherwise, which is what we wanted. $\blacksquare$

We now return to talking about $\zeta$ integrals. For this, we need to fix a $\psi$ for our Fourier transforms, and again, instead of choosing the standard, we choose convenient and force $\psi$ to have conductor $\mf p^0=\mathcal O.$ We can find such a $\psi$ by starting with any nontrivial character $\psi_0$ of conductor $\mf p^\bullet$ and then setting $\psi(x):=\psi_0\left(\varpi^\bullet x\right).$

Instead of dealing with all characters at once in a single proposition, we split up the work a bit. Fix $\chi:F^\times\to\CC^\times$ our character, and we can again decompose $\chi(x):=\eta(x)|x|^s$ for some $\eta$ and $s\in\CC.$ Explicitly, set $\eta:=\chi|_{\mathcal O^\times}$ and then extended to $F^\times$ by $\chi(\varpi)=1$; then $\chi/\eta$ is unramified and equals $|\bullet|^s$ for some $s\in\CC.$

I quickly remark that I think $\eta$ is unitary because $\chi|_{\mathcal O^\times}$ is a character on the compact group $\mathcal O^\times$ and therefore unitary free. Then extending $\eta$ by $\eta(\varpi)=1$ gives $|\eta(x)|=1$ for any $x\in F^\times.$ However, we will not need to know that $\eta$ is unitary anywhere.

This is a straight computation; it doesn't even use the Gauss sums we built. Set our usual Gaussian $f:=1_{\mathcal O}.$ Then\[\zeta(f,\chi)=\int_{F^\times}f(x)\chi(x)\,d^\times x=\int_{\mathcal O\setminus\{0\}}\eta(x)|x|^s\,d^\times x.\]Now, $\eta$ has conductor $1+\mf p^0,$ so it is trivial on all $\mathcal O\setminus\{0\}.$ Thus, this is really an integral over $|x|^s,$ which turns into a geometric series as usual. We see\[\zeta(f,\chi)=\sum_{k=0}^\infty\int_{|x|=q^{-k}}|x|^s\,d^\times x=\sum_{k=0}^\infty q^{-ks}\int_{|x|=q^{-k}}d^\times x.\]The multiplicative measure lets us multiply the inner integral by a uniformizer $\varpi$ until we are integrating over $|x|=1,$ where $d^\times x=dx/|x|=dx.$ Then this all collapses to\[\zeta(f,\chi)=\left(\sum_{k=0}^\infty q^{-ks}\right)\int_{\mathcal O^\times}dx=\frac1{1-q^{-s}}\int_{\mathcal O^\times}dx.\]We could evaluate $\int_{\mathcal O^\times}dx=\frac{q-1}q$ as consisting of $q-1$ of the $q=[\mathcal O:\mf p]$ cosets, but it doesn't really matter. To compare, $\chi(\varpi)=\eta(\varpi)|\varpi|^{-s}=q^{-s},$ so $L(\chi)=\frac1{1-\chi(\varpi)}=\frac1{1-q^{-s}}.$ Thus,\[\frac{\zeta(f,\chi)}{L(\chi)}=\int_{\mathcal O^\times}dx\]is just some constant. Further, $\zeta(f,\chi)$ is nonzero throughout here because $\int_{\mathcal O^\times}dx\ne0,$ and $L(\chi)$ has no zeroes.

As usual, our next step is to compute the Fourier transform of $f=1_{\mathcal O}.$ We have\[\hat f(y)=\int_Ff(x)\psi(xy)\,dx=\int_{\mathcal O}\psi(xy)\,dx.\]Note that $x\mapsto\psi(xy)$ is a trivial character if and only if $\mathcal Oy\subseteq\mathcal O$ because $\mathcal O$ is the conductor of $\psi.$ (Here we see the convenience of our $\psi.$) Certainly $y\in\mathcal O$ implies $\mathcal Oy\subseteq\mathcal O,$ but $y\not\in\mathcal O$ implies $1y\in\mathcal Oy\not\subseteq\mathcal O,$ so this condition is indicating $y\in\mathcal O.$ The point is that\[\hat f(y)=1_{y\in\mathcal O}\int_{\mathcal O}dx=1_{y\in\mathcal O}\]by orthogonality of characters. In particular, our $f$ is self-dual!

Thus, noting that $\chi^\vee(x)=\eta^{-1}(x)|x|^{1-s}$ also gives its $\eta$ part of conductor $\mf p^0,$ we see that the same argument given above establishes\[\frac{\zeta(\hat f,\chi^\vee)}{L(\chi^\vee)}=\int_{\mathcal O^\times}dx.\]Thus, we have the equality claimed in the proposition, and $\zeta(\hat f,\chi^\vee)$ doesn't vanish for the same reasons that $\zeta(f,\chi)$ didn't vanish. So we are done. $\blacksquare$

The harder case is when the conductor of $\eta$ is smaller. We have the following.

This computation is a bit more involved. This time we set $f=1_{1+\mf p^n}$ to make the $\zeta$ integral easier. Indeed, we have\[\zeta(f,\chi)=\int_{F^\times}f(x)\chi(x)\,d^\times x=\int_{1+\mf p^n}\eta(x)|x|^s\,d^\times x.\]By hypothesis, $\eta$ is trivial on $1+\mf p^n,$ so this integral is just over $|x|^s.$ Because $|x|=1$ over $1+\mf p^n,$ we see $|x|^s=1$ and $d^\times x=dx/|x|=dx,$ so\[\zeta(f,\chi)=\int_{1+\mf p^n}dx=q^{-n}\]because, as discussed earlier, $1+\mf p^n$ is just one of $\left[\mathcal O:\mf p^n\right]=q^n$ cosets whose measures sum to $1.$ We see that this is indeed nonzero for any $\chi.$

We now have to compute the Fourier transform of $f.$ We find\[\hat f(y)=\int_Ff(x)\psi(xy)\,dx=\int_{1+\mf p^n}\psi(xy)\,dx.\]We can shift this integral by taking $x\mapsto x-1$ to be over $\mf p^n.$ Then the integrand is $\psi((x+1)y)=\psi(xy+y)=\psi(y)\psi(xy)$ because $\psi$ is an additive character. Thus, this is\[\hat f(y)=\psi(y)\int_{\mf p^n}\psi(xy)\,dx.\]Now we emulate the argument from the previous case to evaluate this integral. If $y\in\mf p^{-n},$ then $xy\in\mathcal O$ always, trivializing the integral. But if $y\not\in\mf p^{-n},$ then $y\varpi^n\not\in\mathcal O,$ meaning that the we are integrating a nontrivial character, causing it to vanish. Thus, this is\[\hat f(y)=\psi(y)\int_{\mf p^n}dx\cdot1_{y\in\mf p^{-n}}=\frac{\psi(y)}{q^n}\cdot1_{y\in\mf p^{-n}}.\]This is good enough for our Fourier transform.

With $\hat f$ in hand, we turn to $\zeta(\hat f,\chi^\vee).$ This time we actually have to do some computation, and only now will Gauss sums appear. Note $\chi^\vee(x)=\eta(x)^{-1}|x|^{1-s},$ so\[\zeta(\hat f,\chi^\vee)=\int_{F^\times}\hat f(x)\chi^\vee(x)\,d^\times x=q^{-n}\int_{\mf p^{-n}\setminus\{0\}}\psi(x)\eta(x)^{-1}|x|^{1-s}\,d^\times x.\]We promised Gauss sums would appear, so we would like to make this integral more like a Gauss sum. To start, we need to be integrating over a multipicative subgroup, so we partition the integral by\[\zeta(\hat f,\chi^\vee)=q^{-n}\sum_{k=-n}^\infty\int_{|x|=q^{-k}}\psi(x)\eta(x)^{-1}|x|^{1-s}\,d^\times x.\]At this point, we use our multiplicative measure to shift the integrals to be over $\mathcal O^\times.$ This amounts to taking $x$ with $|x|=q^{-k}$ to $x\varpi^{-k}.$ Note that $\eta(\varpi)=1,$ so the $\eta$ term remains intact, and $|x|^{1-s}=q^{-k(1-s)}$ we may just factor out, collapsing everything to\[\zeta(\hat f,\chi^\vee)=q^{-n}\sum_{k=-n}^\infty q^{-k(1-s)}\int_{\mathcal O^\times}\eta(x)^{-1}\psi\left(\varpi^kx\right)\,d^\times x.\]And finally, this integral is the Gauss sum $g\left(\eta^{-1},x\mapsto\psi\left(\varpi^kx\right)\right).$ Our multiplicative character $\eta^{-1}$ still has conductor $1+\mf p^n,$ and our additive character has conductor $\varpi^{-k}\mf p^0=\mf p^{-k}.$ Using our lemma on Gauss sums, then, this term vanishes except for when $k=-n$ causes the conductors to match. When the dust settles, we have\[\zeta(\hat f,\chi^\vee)=q^{-n}\cdot q^{-n(1-s)}\cdot g\left(\eta^{-1},x\mapsto\psi\left(\varpi^{-n}x\right)\right).\]Each term of this product is always nonzero—the Gauss sum has magnitude $q^{-n}$ and is therefore nonzero—so indeed, $\zeta(\hat f,\chi^\vee)$ is again nonzero. It only remains to compute $\varepsilon,$ which comes out to be\[\frac{\zeta(\hat f,\chi^\vee)}{\zeta(f,\chi)}=q^{-n(1-s)}g\left(\eta^{-1},x\mapsto\psi\left(\varpi^{-n}x\right)\right)\]because $\chi$ being ramified implies $L(\chi)=L(\chi^\vee)=1.$ This finishes the proof of the proposition. $\blacksquare$

We are now ready for payoff. Recall that for Shwartz-Bruhat functions $f,$ we know $\zeta(f,\chi)$ converges for $\chi$ of positive exponent $\sigma.$ This means that $\chi^\vee$ has exponent $1-\sigma,$ so $\zeta(f,\chi^\vee)$ will converge for exponents $\sigma \lt 1.$ This motivates the following statement.

For this we simply have to do the manipulation suggested at the beginning. Our character $\chi$ has some unitary part $\eta,$ and the above long casework lets us pick out some $f$ for which\[\frac{\zeta(\hat f,\chi^\vee)}{L(\chi^\vee)}=\varepsilon(\chi,\psi,dx)\frac{\zeta(f,\chi)}{L(\chi)}\]always holds, and the $\zeta$ integrals are always nonzero. Well, both $\zeta$ integrals are defined over $\op{Re}(s)\in(0,1),$ so the functional equation holds there. To convert this to $g,$ we do as suggested and write\[\frac{\zeta(\hat g,\chi^\vee)}{L(\chi^\vee)}=\frac{\zeta(\hat g,\chi^\vee)}{\zeta(\hat f,\chi^\vee)}\cdot\frac{\zeta(\hat f,\chi^\vee)}{L(\chi^\vee)}=\frac{\zeta(g,\chi)}{\zeta(f,\chi)}\cdot\varepsilon\frac{\zeta(f,\chi)}{L(\chi)}=\varepsilon\frac{\zeta(g,\chi)}{L(\chi)},\]which finishes the proof. $\blacksquare$

And this lets provide an analytic continuation for the $\zeta$ integrals.

Indeed, for $\op{Re}(s) \gt 0,$ we already have convergence of $\zeta(f,\chi).$ Then for $\op{Re}(s) \lt 1,$ we define\[\zeta(f,\chi):=L(\chi)\cdot\varepsilon(\chi,\psi,dx)\frac{\zeta(\hat f,\chi^\vee)}{L(\chi^\vee)}\]provides a funtional equation to extend. Indeed, the regions of our definition for $\op{Re}(s) \gt 0$ and $\op{Re}(s) \lt 1$ agree on the open set $\op{Re}(s)\in(0,1),$ so in fact their meromorphic continuations must agree everywhere. This establishes what we wanted. $\blacksquare$

As an end remark, we note that our $\varepsilon$s tended to be well-behaved. Most of the time $\varepsilon$ didn't even depend on $s,$ with the exception of the ramified characters in nonarchimedean fields. Nevertheless, in all cases are $\varepsilon$ at least looks like $ae^{bs}$ for some $a,b\in\CC.$ For example, in the last case, $b=n\log q,$ I think.

Nevertheless, there will turn out to be few cases of the last case: if we fix a global (say, number) field $K,$ then we note that for all unramified primes $\mf p,$ the different $\mathcal D_\mf p$ vanishes, so the stated measure and $\psi$ are accurate are the standard ones. And our characters will be trivial over all but finitely many $\mf p,$ defaulting us to the easy nonarchimedean case. So that case isn't too worrisome.