## Thursday, February 9, 2012

### Volume 1


Page 34, Exercise 1.12.c first line: $\F_p$'' should be $\F_q$''

Page 65, Theorem 2.4.15 in the functional equation, the $L$ function in the denominator of the left-hand side should be evaluated at $1-s,$ rather than $s.$ Thus the functional equation is
$$\frac{Z_v(1-s, \, \widehat \Phi, \, \chi^{-1})}{L_v(1-s, \chi^{-1})} = \epsilon_v(s, \chi) \frac{Z_v(s, \Phi, \chi)}{L_v(s, \chi)}, \qquad \tx{ not } \qquad \frac{Z_v(1-s, \, \widehat \Phi, \, \chi^{-1})}{L_v(s, \chi^{-1})} = \epsilon_v(s, \chi) \frac{Z_v(s, \Phi, \chi)}{L_v(s, \chi)}.$$

Page 80, Proposition 3.3.7 is not correct. Niccolo' Ronchetti was particularly helpful with the details of this erratum. First, note that reduction modulo $N$ defines a bijection between ${\Gm_0(N)\bs SL(2,\Z)}$ and ${B(\Z/N\Z)\bs SL(2,\Z/N\Z)},$ where $B(\Z/N\Z)$ is the subgroup of $SL(2,Z/N\Z)$ consisting of upper triangular elements. If $N=\prod_p p^{e_p}$ then the isomorphism $\Z/N\Z \to \prod_p \Z/p^{e_p} \Z$ induces a bijection
$$B(\Z/N\Z)\bs SL(2,\Z/N\Z) \to \prod_p B(\Z/p^{e_p}\Z)\bs SL(2,\Z/p^{e_p}\Z)$$
From this, it is clear that if $n(N)$ is the number of cosets of $\Gm_0(N)$ in $SL(2,\Z),$ then
$$n\left( \prod_p p^{e_p} \right) = \prod_ p n\left( p^{e_p}\right).$$
However, in the case $N=12$ proposition 3.3.7 produces only $22$ coset representatives for ${\Gm_0(12)\bs SL(2,\Z)}.$
$$\bpm *&*\\ 1&1 \epm, \bpm *&*\\ 2&1 \epm, \bpm *&*\\ 3&1 \epm, \bpm *&*\\ 4&1 \epm, \bpm *&*\\ 5&1 \epm, \bpm *&*\\ 6&1 \epm, \bpm *&*\\ 7&1 \epm, \bpm *&*\\ 8&1 \epm, \bpm *&*\\ 9&1 \epm, \bpm *&*\\ 10&1 \epm, \bpm *&*\\ 11&1 \epm,$$
$$\bpm *&*\\ 12&1 \epm, \bpm *&*\\ 1&2 \epm, \bpm *&*\\ 3&2 \epm, \bpm *&*\\ 5&2 \epm, \bpm *&*\\ 1&3 \epm, \bpm *&*\\ 2&3 \epm, \bpm *&*\\ 4&3 \epm, \bpm *&*\\ 1&4 \epm, \bpm *&*\\ 3&4 \epm, \bpm *&*\\ 1&6 \epm, \bpm *&*\\ 1&12 \epm.$$
However, $\Gm_0(12)\bs SL(2,\Z)$ actually has $24$ cosets. Certainly, each coset contains a representative of the form $\bspm *&*\\ c&d \espm$ with $d\mid N,$ and $\bspm *&*\\ c_1&d \espm$ is in the same coset as $\bspm *&*\\ c_2&d \espm$ if and only if ${c_1 \equiv mc_2 \pmod{N}}$ for some $m \equiv 1 \pmod{N/d}.$ Thus we are reduced to studying orbits of $$\{ m \in (\Z/N\Z)^\times\mid m \equiv 1 \pmod{N/d}\}$$
on $$\{ c \in \Z/N\Z \mid \gcd(c,d) =1\}.$$ For divisors of $12,$ if $d \ne 3,4,$ then each orbit has a unique representative between $1$ and $N/d.$ If ${d=3},$ then one is considering the action of $\{1,5\}$ on $\{1,2,4,5,7,8,10,11 \}.$ (With the action being defined by multiplication mod $12.$) The orbits are $\{1,5\}, \{ 2,10\}, \{4, 8\},$ and $\{7,11\}.$ The last has no representative between $1$ and $N/d,$ which is $4$ in this case.

Likewise, if $d =4$ one considers the action of $\{1,7\}$ on $\{1,3,5,7,9,11\}.$ The orbits are $\{1,7\}, \{3,9\}, \{5, 11\},$ and the last has no representative between $1$ and $N/d=3.$

Page 91, Proposition 3.9.7 is not correct as stated. Specifically, there exist elements $f$ of ${\c A^*_{k,\chi}(\Gm_0(N))}$ such that $L_k.f$ and/or $R_k.f$ are not of moderate growth. For example one might define $f$ which is supported on ${\{ x+iy\mid |x| \le 13, \; y \ge 2\}}$ and equal to $g(y)$ on ${\{ x+iy\mid |x| \le 14, \; y \ge 3\}},$ where $g(y)$ is a function such as $\sin(e^y)$ which is of moderate growth in $y$ but has a derivative which is not of moderate growth. The support of this function is contained in the interior of a fundamental domain for $SL(2, \Z),$ and hence for $\Gm_0(N)$ for any $N.$ It follows that one can extend it to an elementof $\c A^*_{k, \chi}(\Gm_0(N),$ for any $k, \chi$ and $N,$ simply by taking the automorphic relations to be the definition outside of the fundamental domain. In the region where $f(z) = g(y),$ one has $$[R_k.f](z) = yg'(y)-\frac k2 g(y)\qquad [L_k.f](z) = yg'(y)+\frac k2 g(y),$$so neither is of moderate growth.

There are (at least) two ways to fix this. The simplest is to remove the growth condition. If $f$ is a smooth function which satisfies the weight k automorphic relations (3.5.1) on p. 82, then ${R_k.f}$ (resp. ${L_k.f}$) will satisfy the analogous relations in weight ${k+2}$ (resp. ${k-2}$).

The second option is to impose the eigenfunction condition earlier. If $f$ is an automorphic function as in 3.5.2, $and$ also an eigenfunction of the weight $k$ Laplacian, then one can show that $R_k.f$ and $L_k.f$ are also of moderate growth. As far as we know this is not remotely obvious. We don't know a proof which does not rely on the theorem of Harish-Chandra given in 5.1.3.

Page 92, in the formula at the top of the page. The denominator of the second fraction
should have a bar over the $z$ and the denominator of the third fraction should have absolute value symbols in place of parentheses. The corrected version is:
$$(w - \bar w) \Big|_{w = \frac{az+b}{cz+d}} \; = \; \frac{az+b}{cz+d} - \frac{a\bar z +b}{c\bar z+d} \; = \; \frac{z - \bar z}{|cz+d|^2}.$$

Page 99, Chapter 3 Exercise 2(c) is not true as stated. (As a counterexample, take $G=\R^\times$ acting on $X = \R$ by multiplication. The orbits are $\{ 0\}$ and $\R^\times$ and the latter is certainly dense. Define $\psi(g,x)$ to be $g$ for all $g\in \R^\times$ and $x \in \R.$ And take $f(x) = x$ for all $x\in \R.$) In order to be true, the problem should include the requirement that $|\psi(g,x)|=1$ for all $g \in G$ and $x \in X.$

Page 115, Proof of Lemma 4.5.4. The statement that $\delta(E_{i,j})$ is $\ptl{}{g_{i,j}}$ is simply false. The correct formula is
$$\delta(E_{i,j}) = \sum_{k=1}^2 g_{k,i} \ptl{}{g_{k,j}}.$$
Let
$$d:= (d_1, d_2, d_3, d_4), \qquad D_d := \delta\left( E_{1,1}^{d_1} \circ E_{1,2}^{d_2} \circ E_{2,1}^{d_3} \circ E_{2,2}^{d_4}\right).$$
In order to fill the gap in the proof of lemma 4.5.4, we must show that $\{ D_d\mid d \in \Z_{\ge 0}^4\}$ is a linearly independent set of differential operators.

To do so, consider the action of $D_d$ on $e^{g_{1,1}+g_{1,2}+g_{2,1}-g_{2,2}}.$ Clearly,
$$D_d e^{g_{1,1}+g_{1,2}+g_{2,1}-g_{2,2}} = P_d(g) e^{g_{1,1}+g_{1,2}+g_{2,1}-g_{2,2}},$$
where $P_d(g)$ is a polynomial in the entries of $g$ such that
$$P_d(g) = (g_{1,1}+g_{2,1})^{d_1} (g_{1,1}-g_{2,1})^{d_2}(g_{1,2}+g_{2,2})^{d_3}(g_{1,2}-g_{2,2})^{d_4} + \tx{ lower order terms.}$$
Suppose that the differential operators ${\{D_d\mid d \in \Z_{\ge 0}^4\}}$ are linearly dependent. Then the polynomials ${\{P_d\mid d \in \Z_{\ge 0}^4\}}$ are linearly dependent. Write ${||d|| := d_1+d_2+d_3+d_4}.$ And suppose that
$$\sum_{d \in S} c_d P_d = 0,$$
for some finite set $S,$ and some coefficients $c_d \in \C,$ which are all nonzero. In order for a nonzero linear combination of the polynomials ${\{P_d\mid d \in \Z_{\ge 0}^4\}}$ to cancel, there must be cancellation among the terms of highest order which appear.

Write $M = \max_{d \in S} ||d||.$ Then
\sum_{d \in S, ||d||=M}
c_d (g_{1,1}+g_{2,1})^{d_1}

Not that if this identity holds for all $g \in GL(2,\R),$ then it forces the right hand side to be the zero polynomial, and so the identity holds for all $g \in \Mat_{2\times 2}(\R).$

Set $w=g_{1,1}+g_{1,2}, x= g_{1,1}-g_{1,2}, y = g_{2,1}+g_{2,2}$ and $z=g_{2,1}-g_{2,2}.$ Suppose (1) holds for all $g\in GL_2(\R).$ Then it holds for all $g \in \Mat_{2\times 2}(\R).$ But then
$$\sum_{d \in S, ||d||=M} c_d w^{d_1} x^{d_2}y^{d_3}z^{d_4}=0, \qquad ( \forall w,x,y,z \in \R)$$
And then $c_d = 0$ for all $d\in S, ||d||=M,$ a contradiction.

Page 119, the Remark promises exercises showing that a "slight weakening of (2) follows from (3) and (4)." This promise was not kept, and on closer inspection, the statement that we had in mind relies on (1) as well as (3) and (4). Here is what we had in mind. Let us say that a function $\phi: GL(2, \A_\Q) \to \C$ is $Z(GL(2, \A_\Q))$-finite if the space spanned by translates of $\phi$ by elements of the center of $GL(2, \A_\Q)$ is finite dimensional. Then the assertion is that any function which satisfies (1), (3) and (4) is $Z(GL(2, \A_\Q))$-finite. Note that condition (2) can be reformulated as saying that the space spanned by translates of $\phi$ by elements of the center of $GL(2, \A_\Q)$ is one dimensional. Thus, $Z(GL(2, \A_\Q))$-finiteness is indeed a "slight weakening of (2)". Now, how does one prove that a smooth function $\phi: GL(2, \A_\Q) \to \C$ which satisfies (1), (3) and (4) is $Z(GL(2, \A_\Q))$-finite? Well, by strong approximation, proposition 1.4.6, we can express $Z(GL(2, \A_\Q))$ as the product of $Z(GL(2, \Q)),$ $Z(GL(2, \A_Q)) \cap K$ and $(0, \infty)$, where $(0, \infty)$ is embedded into $Z(GL(2, \A_\Q))$ as scalar matrices at the place infinity: $$\iota_\infty(x) := i_\infty\left(\bpm x&0 \\ 0 & x \epm\right), \qquad (x \in \R, \ x > 0 ).$$ Then condition (1) implies that all translates of $\phi$ by elements of $Z(GL(2, \Q))$ are equal to $\phi$ itself. Condition (3) implies that the translates by elements of $Z(GL(2, \A_Q)) \cap K$ remain in a finite dimensional subspace. If we can show that the translates of $\phi$ by $\iota_\infty(x), \ (x \in \R, \ x>0)$ remain in a finite dimensional vector space, we will be done. Condition (4) implies that $$D_Z^n .\phi, \qquad ( n \in \Z, \ n \ge 0 )$$ spans a finite dimension vector space. Here $D_Z$ is defined as in 4.5.1, taking $Z$ to be the $2 \times 2$ identity matrix, so that $D_Z$ is in the center of $U(\f{gl}(2, \C)).$ Now, if we choose a basis for our finite dimensional vector space, then $D_Z^n$ will act on it by some matrix. We may arrange for the matrix to be in Jordan canonical form, and then our general assertion follows from the special case when there is only one block. In that case there is a positive integer $n$ and a complex number $\lambda$ such that $(D_Z-\lambda)^n.\phi =0.$ Now, $$D_Z.\phi(g) = \frac{d}{dt} \phi( g \cdot \iota_\infty( e^t))\big|_{t=0}, \qquad ( g \in GL_(2, \A_\Q)).$$ So, we might define a differential operator $D: C^\infty((0, \infty)) \to C^\infty((0, \infty))$ by $D.f(x) =\frac{d}{dt} f(x e^t)\big|_{t=0}.$ Then the space of solutions of the equation $(D-\lambda)^n.f = 0$ is spanned by $\{ x^\lambda (\log x)^j : 0 \le j < n\}.$ It follows that $$\phi(g \iota_\infty(x)) = \sum_{j=0}^{n-1} c_j(g) x^\lambda (\log x)^j,$$ for some functions $c_j: GL(2, \A_\Q) \to \C.$ Thus every translate of $\phi$ by an element of the form $\iota_\infty(x)$ remains in the finite dimensional vector space spanned by these functions $c_j, 0 \le j < n,$ and we are done.
Page 150, Exercise 4.17(b) The statement is not correct. The problem is with $Z(U(\f{gl}_2(\C)))$-finiteness. A correct version would be
Let $f: \mathbb{R}^+ \to \mathbb{C}$ be a smooth function, and consider the function $\phi: GL(2, \mathbb{A}_\mathbb{Q}) \to \mathbb{C}$ given by $$\phi(g) = f(|\det(g)|_{\mathbb{A}}).$$ Characterize those functions $f$ for which $\phi$ satisfies (1) and (3)-(5) of Definition~4.7.6 for an adelic automorphic form. Then determine which of your functions have central character and which do not. A solution would be as follows: conditions (1) and (3) are trivial for any $f.$ To analyze condition (4), write $\phi(g) = f( c(g) )$, where $c(g) = |\det(g)|_\mathbb{A}$. Then $D_\alpha .\phi(g) = \operatorname{Tr}(\alpha) c(g) f'(c(g))$. Taking $\alpha = Z$, we get a subspace containing $f(c(g))$ and $c(g) f'(c(g)).$ A composition $D_{\alpha}\circ D_{\beta}$ acts by $$D_{\alpha} \circ D_{\beta}.\phi(g) = \operatorname{Tr}(\alpha)\operatorname{Tr}(\beta) [ c(g) f'(c(g)) + c(g)^2 f''(c(g)) ].$$ It follows by induction that if $f$ is $Z(U({\frak g}))$ finite, then y = f(x) is a solution to the differential equation $$x^n y^(n) + a_{n-1} x^{n-1} y^(n-1) + ... + a_1 x y' + a_0 y = 0$$ for some constants $a_0, \ldots, a_{n-1}$. The solutions to this equation are linear combinations of functions of the form $y = x^a (\log x)^b$, where a is a complex number and b < n is a nonnegative integer. Note that if $f$ is any function of this type, then $\phi$ also satisfies (5). On the other hand, $\phi$ has a central character if and only if $f(x)$ is a scalar multiple of $x^a$ for some complex number $a.$
Page 177, proof of theorem 5.5.15 There is a reference to theorem 5.4.1. But no such theorem exists. The correct reference is lemma 5.1.7
Page 188-89, proof of theorem 6.1.11, definitions of $\c S$ and $\c S(\chi)$ are incompatible. More precisely, the group $T_\tau$ is not compact: it is only compact mod center. Now, if $f:GL(2, \Q_p) \to \C$ is a function satisfying
$$f(t_1gt_2) = \chi(t_1t_2) f(g), \qquad ( \forall t_1, t_2 \in T_\tau, \;\; g \in GL(2, \Q_p)),$$ for some character $\chi$ of $T_\tau,$ then the support of $f$ is certainly a union of $T_\tau$-double cosets-- and hence can be compact only if it is empty. This was pointed out to us by Omer Offen.

To correct this, let $Z$ denote the center of $GL(2, \Q_p),$ and fix a character $\omega$ of $Z.$ Define $\c S_\omega$ as the space of smooth functions $GL(2, \Q_p)$ such that

(1) $f( zg) = \omega(z) f(g)$ for all $g \in GL(2, \Q_p)$ and all $z\in Z,$ and

(2) $f$ is compactly supported modulo the center, (that is, the support of $f$ is contained
$$\{ cz \mid z \in Z, \;\; c \in C\}$$
for some compact set $C \subset GL(2, \Q_p)$).

Let $\omega_\pi$ be the central character of $\pi.$ Then for each $v \in V,$ and $f \in\c S_{\omega_\pi^{-1}}$ the function ${g \mapsto f(g) \cdot \pi(g).v}$ is invariant by $Z.$ Hence for $f \in \c S_{\omega_\pi^{-1}}$ one may define an operator $\pi(f)$ by integrating of ${GL(2, \Q_p)/Z.}$ That is, change the definition of $\pi(f)$ given in (6.1.12) to be
$$\pi(f)\,.\,v := \int_{GL(2, \Q_p)/Z} \pi(g)\,.\,v \;\; f(g) \, dg, \qquad ( \forall v \in V, \;\; f \in \c S_{\omega_\pi^{-1}}).$$
One needs to modify Claim 1 as well as its proof, so that they apply to $\c S_{\omega_\pi^{-1}},$ rather than the original space $\c S.$ This is not difficult. Take $v \in V.$ As stated in the proof of Claim 1, one may choose a compact set $U$ such that $\pi(u)\,.\,v = v$ for all $u \in U.$ One may then define an element $f$ of $\c S_{\omega_\pi^{-1}}$ by
$$f( g) = \begin{cases}\omega_\pi^{-1}(z), & g = u z, \qquad u \in U,\; z \in Z,\\ 0& g \text{ not of this form.} \end{cases}$$
Then $\pi(f)\,.\,v$ is equal to $v$ times the volume of $U.$ With this in hand it is easy to modify the proof of Claim 1 for $\c S$ so that it works for $\c S_{\omega_\pi^{-1}}.$

Next,if $\chi$ is a character of $T_\tau$ then the restriction of $\chi$ to $Z$ is some character $\omega$ of $Z.$ Define $\c S(\chi)$ as
$$\c S(\chi) := \left\{f \in \c S_{\omega} \mid f(t_1gt_2) = \chi(t_1 t_2) f(g), \; (t_1, t_2 \in T_\tau, \; g \in GL(2, \Bbb Q_p))\right\}.$$
There are two differences between this definition and the original definition given on p.189 in the book. The first is that one has $f \in \c S_{\omega},$ instead of $f \in \c S.$ The second is that one has $\chi(t_1 t_2)$ instead of $\chi(t_1 t_2)^{-1}.$ If the restriction of $\chi$ to $Z$ is $\omega_\pi,$ then $\c S(\chi^{-1}) \subset \c S_{\omega_\pi^{-1}},$ and so we have an operator $\pi(f): V \to V$ for each $f \in \c S(\chi),$ defined as above.

Likewise, in equation (6.1.14), the integral which defines the convolution should be taken over $GL(2, \Q_p)/Z.$ To see this, take $f_1, f_2 \in \c S(\chi)$ for some $\chi$ and $g \in GL(2, \Q_p),$ and check that the function $F(h) := f_1(h^{-1} g) f_2(h)$ satisfies $F(zh) = F(h), \;(\forall h \in GL(2, \Q_p), \; z \in Z).$

page 188, penultimate formula should read $$\frak t^{-1} \begin{pmatrix} a & b\tau\\b & a\end{pmatrix} \frak t \; = \;\begin{pmatrix} a+b\sqrt{\tau} & 0\\0& a-b\sqrt{\tau}\end{pmatrix},$$ rather than $$\frak t^{-1} \begin{pmatrix}a & b\tau\\b & a\end{pmatrix} \frak t \; = \;\begin{pmatrix} a+b\tau & 0\\0& a-b\sqrt{\tau}\end{pmatrix}.$$ (That is, there is a missing $\sqrt{\phantom{\tau}}.$)
page 189, Claim 2 and definition of $\c S(\chi)$ are not consistent. The proof of Claim 2 is essentially correct, but it follows from a change of variables that the function $F$ which is constructed satisfies $F( t_1 g t_2) = \chi(t_1t_2)^{-1} F(g)$ for all $t_1, t_2 \in T_\tau$ and $g \in GL(2, \Q_p).$ According to the definition of $\c S( \chi)$ given on p. 189, this would put $F$ in $\c S( \chi),$ not $\c S(\chi^{-1}).$ This can be corrected by modifying the definition of $\c S(\chi)$ according to the previous correction. Likewise, the proof that $\c S(\chi^{-1})$ is closed under convolution given on p.190 implicitly uses the definition of $\c S(\chi)$ given above, and not the one which appears on p.189.

page 190, line -2 states that "${GL(2, \Z_p)}$ normalizes every $K'$." Since $K'$ is an arbitrary compact open subgroup of ${GL(2, \Z_p)},$ this is not true. Instead, we should require that $K'$ be normal in ${GL(2, \Z_p)}$ when $K'$ is introduced two lines earlier. There is no harm in this, since every open subgroup of ${GL(2, \Z_p)}$ contains the subgroup $K_n$ introduced on p. 183 in (6.1.3) for some positive integer $n,$ and since the subgroup $K_n$ is normal for every $n.$
Page 234, proof of proposition 6.10.4, equation (6.10.5) should read $$\Big\{ \phi: \Q_p^\times \to \C\;\Big|\; \phi(y) =c\cdot \omega_2(y) \, |y|_p^{s_2+1} +f(y), \quad \left(c\in \C, \; f \in S(\Q_p^\times)\right)\Big\},$$ rather than $$\Big\{ \phi: GL(2, \Q_p)\to \C\;\Big|\; \phi(y) =c\cdot \omega_2(y) \, |y|_p^{s_2+1} +f(y), \quad \left(c\in \C, \; f \in S(\Q_p^\times)\right)\Big\}.$$ (The domain is $\Q_p^\times,$ not $GL(2, \Q_p).$)

Page 234, three lines after equation (6.10.5), there is a reference to theorem 6.6.15,'' which does not exist. The correct reference is theorem 6.8.11.
Page 318-319, Proposition 8.10.3 and its proof It says that the measure on the right hand side of (8.10.4) assigns $K_1$ a measure of $p^{-2}.$ But computing the measure of $K_1$ using the product measure $$\frac{d\alpha\ d\beta\ d\gamma \ d\delta}{|\alpha \delta - \beta \gamma|^2}$$ would mean computing $$\int_{1+p \Z_p } \int_{p \Z_p} \int_{p \Z_p} \int_{1+p \Z_p} 1 \ d\alpha\ d\beta\ d\gamma \ d\delta,$$ and the answer is $p^{-4},$ not $p^{-2}.$ As a side effect, the scalar which relates the two measures should be $$\frac{p^3}{(p^2-1)(p-1)}, \qquad \text{ not }\qquad \frac{p}{(p^2-1)(p-1)}.$$ Page 336, Theorem 8.11.18 Is clearly false as stated: one may easily construct two piecewise continuous functions which differ only on a set of measure zero. For example, altering the value of a continuous function at a single point produces a function which is still piecewise continuous. We may say two functions are equivalent if they differ on a set of measure zero. Then the Mellin transform depends only on the equivalence class. The question, then is which element of the class is given by the inverse Mellin transform of the Mellin transform?

By a piecewise continuous function, we mean a function $f:[0,\infty) \to \C$ such that $\lim_{x\to a^+}f(x)$ and $f(a)$ are defined for all $a \in [0,\infty),$ and $\lim_{x \to a^-}f(x)$ is defined for all $a \in (0,\infty),$ and $$\lim_{x \to a^-}f(x)= \lim_{x \to a^+}f(x)=f(a)$$ outside of a discrete set. Any such function is equivalent to one which satisfies $$f(a) = \frac 12 \left(\lim_{x \to a^-}f(x)+ \lim_{x \to a^+}f(x)\right).$$ It is this equivalence-class member to which the inverse Mellin transform converges. See, for example, section 1.29 of Titchmarsh's Theory of Fourier Integrals.

For our purposes it is enough to know that Mellin inversion holds as stated for continuous functions, including Schwarz functions.

Page 416, Exercise 10.19 ought to have the additional assumption that the representation $V$ is irreducible. Also the requirement that $G$ have an open maximal compact subgroup does not appear to be needed. A hint for this problem is to consider the proofs of lemmas 8.5.11 or 8.7.3, as opposed to that of 6.1.8 (where admissibility is not assumed and we have to work harder). The statement also appears as proposition 4.2.4 in Bump's Automorphic forms and representations.

Page 427 When the substitution $g \mapsto \xi \tau h_1$ is made, it is assumed that $f_1$ is independent of $\tau.$ This is equivalent to assuming that the central character of our cuspidal representation is the idelic lift of some Dirichlet character. We saw in Chapter two that every unitary Hecke character is of the form $a \mapsto \chi_{\on{idelic}}(a)|a|^{i\nu}$ for some Dirichlet character $\chi$ and some real number $\nu.$ Thus the proof as written only treats the case $\nu = 0.$ (Thanks to Samuel Mundy for pointing this out.) The extension to general $\nu$ is fairly straightforward: one just gets a vertical shift in the $s$ variable.

The next series of corrections was pointed out to us by Gergely Harcos. Page 451-52, starting at the top of p. 451, we use the formula $$L_p(s, \chi_i) = (1-\chi_i(p) p^{-s}),\qquad (i=1,2)$$ which is only valid if $\chi$ is unramified. In the case when $\chi$ is ramified, then of course $\chi_1$ and $\chi_2$ are also ramified and the correct formula is $$L_p(s, \chi_i) = 1,\qquad (i=1,2).$$ (See (2.4.2) and (2.4.3).)

p. 451, in the middle \begin{aligned} I(s) & := -\chi^{-1}(p)p^{s-\frac12} \epsilon(s, \chi_1) \epsilon(s, \chi_2)\int\limits_{GL(2, \Q_p)} \frac{\Phi(g)\, \beta(g) \, |\det(g)|_p^{s+\frac12} }{ \left(1-\chi(p)p^{-s-\frac12}\right)^{-1} } \; d^\times g\\ & \hskip 23pt = II(s) \; := \;\int\limits_{GL(2, \Q_p)} \frac{\widehat{\Phi}(g)\, \beta\left(g^{-1}\right) \, |\det(g)|_p^{\frac 32-s} }{ \left(1-\chi^{-1}(p)p^{s-\frac32}\right)^{-1} } \; d^\times g.\end{aligned} should be \begin{aligned} I(s) & := -\chi(p)p^{-s+\frac12} \epsilon(s, \chi_1) \epsilon(s, \chi_2)\int\limits_{GL(2, \Q_p)} \frac{\Phi(g)\, \beta(g) \, |\det(g)|_p^{s+\frac12} }{ \left(1-\chi(p)p^{-s-\frac12}\right)^{-1} } \; d^\times g\\ & \hskip 23pt = II(s) \; := \;\int\limits_{GL(2, \Q_p)} \frac{\widehat{\Phi}(g)\, \beta\left(g^{-1}\right) \, |\det(g)|_p^{\frac 32-s} }{ \left(1-\chi^{-1}(p)p^{s-\frac32}\right)^{-1} } \; d^\times g.\end{aligned}. That is, it should be $-\chi(p)p^{-s+\frac12}$ and not $-\chi^{-1}(p)p^{s-\frac12}.$

p.452, definition of $L_p(s, \pi)$ should read $$L_p(s, \pi) = L_p(s, \chi_2) = \begin{cases} \left( 1 - \chi(p) p^{-s - \frac 12}\right)^{-1}, & \chi \text{ unramified}, \\ 1 & \chi \text{ ramified.} \end{cases}$$ p. 452, definition of $\epsilon_p(s, \pi)$ should read $$\epsilon_p(s, \pi) = \begin{cases} -\chi(p) \epsilon(s, \chi_1)\epsilon(s, \chi_2) p^{-s+\frac12}, & \chi \text{ unramified}, \\ \epsilon(s, \chi_1)\epsilon(s, \chi_2), & \chi \text{ ramified.} \end{cases}$$
p. 492, solution to 4.17(b) the equation which is said to give (4) does not, in fact, do so. Indeed, condition (4) only holds if $f$ is of a very special type. See the correction to 4.17(b) itself.