p9, top. It is an isomorphism of riemannian manifolds that is called an isometry, not a morphism.
p12, footnote 10. It was H. Cartan, not E. Cartan, who proved that the group of isometries of a bounded domain has a natural structure of a Lie group; then E. Cartan proved that the group of isometries of a symmetric bounded domain is semisimple (see Borel, Essays …, 2001, IV 6).
p30, proof of 2.14, just above the diagram 26. I write $G\overset{\mathrm{Ad}}{\longrightarrow}\mathfrak{g}{}$ where I should write $G\overset{\mathrm{Ad}}{\longrightarrow}\mathrm{GL}(\mathfrak{g}{})$ (from Bin Du).
p.51, l.5 ... depend on a (from Timo Keller).
p.59, Lemma 5.22 This is misstated: in general, $T(\mathbb{Q}{})$ is not closed in $T(\mathbb{A}_{f})$ (unless $(G,X)$ satisfies SV5) and so $T(\mathbb{Q}{})\backslash T(\mathbb{A}{}_{f})$ is not Hausdorff (hence not compact). The last step of the proof “ An arbitrary torus ...”\ fails when $T(\mathbb{Q}{})\backslash T(\mathbb{A}% {}_{f})$ is not compact. The proof of the finiteness of $T(\mathbb{Q}% {})\backslash T(\mathbb{A}{}_{f})/\nu(K)$ needs to be rewritten. (Bas Edixhoven)
p.67. In the display under SV1, interchange $z/\bar{z}$ and $\bar {z}/z$.
p80, 81. Lucio Guerberoff points out that the uniqueness assertion in Proposition 8.14 fails and that the condition (**) in Theorem 8.17 is inadequate. He writes (slightly edited):
In Theorem 8.17, you say that your condition (**) on the isomorphism $a$ is enough to guarantee that $ah_{A}$ belongs to $X$ ($h_{A}$ being the morphism defining the Hodge structure on $H_{1}(A,\mathbb{Q}{})$). However I believe that this only implies that $ah_{A}$ belongs to the Siegel double space of $(V,\psi)$, but not necessarily to the $G(\mathbb{R}{})$-conjugacy class $X$. More precisely, I'm not sure if I'm missing something in Proposition 8.14. I tried to reproduce all the relevant calculations, comparing with Kottwitz's JAMS paper, and my conclusion is that if $x$ is a morphism from $\mathbb{S}{}$ to $G_{\mathbb{R}{}}$, then $x$ belongs to $X$ if and only if two conditions hold:1) $x$ lies in the Siegel double space of $(V,\psi)$, and
2) (fix one morphism $h$ in $X$) the two $B\otimes_{\mathbb{Q}}\mathbb{C}$ structures on $V\otimes_{\mathbb{Q}}\mathbb{R}$ (one is $x(i)$, other one is $h(i)$) are isomorphic.
Condition 1) only guarantees that they will be $\mathbb{C}$-isomorphic, not necessarily in a $B$-linear way. I don't see how this would be automatically implied from 1). In other words, your statement of Proposition 8.14 suggests that if $x$ and $h$ have target $G_{\mathbb{R}},$ satisfy 1) and are conjugate under $\mathrm{GSp}(\psi)(\mathbb{R})$, then they are also $G(\mathbb{R}% )$-conjugate, which doesn't seem to be the case. (For example, take a unitary group of signature $(r,s)$ over a CM extension $K/\mathbb{Q}{}$, $K$ quadratic imaginary say, and starting with the usual $h(z)=\mathrm{diag}(zI_{r},\bar{z}I_{s})$, consider $h^{\prime}(z)=h(\bar{z})$; then $h^{\prime}$ has target $G_{\mathbb{R}}$, and is obviously on the Siegel double space (to form $\psi$, use a trace zero element in $K$), but it's not $G(\mathbb{R})$-conjugate to $h$ unless $r=s$).
In the same vein, I'm seeing condition (**) as missing something. In the same example, suppose the hermitian space defining the unitary group is $(V,\langle,\rangle)$, and consider $(V,-\langle,\rangle)$, so it has signature $(s,r)$. Neither of the conditions on Theorem 8.17 care about whether you look at $\langle,\rangle$ or $-\langle,\rangle$, but the Shimura variety of $-\langle,\rangle$ should be the complex conjugate of the variety for $\langle,\rangle$.
p100, top line (proof of 11.2). Delete “ therefore ”from “ The map therefore factors through …” --- as Brian Conrad reminded me the group $\mathbb{A}{}_{E,f}^{\times}/E^{\times}$ need not be Hausdorff. In a detailed proof, one replaces $\mathbb{A}{}_{E,f}^{\times}/E^{\times}$ with a quotient $T(\mathbb{A}_{f})/T(\mathbb{Q}{})$, which is Hausdorff. Here $T$ is a certain subtorus of $(\mathbb{G}_{m})_{E/\mathbb{Q}}$. See my notes on Complex Multiplication for the details.
p124, top. Shenghao Sun has pointed out to me that the statement that pro-tori correspond to free $\mathbb{Z}{}$-modules with a continuous action of $\Gamma$ is contradicted on the next page where I show that the character group of $\mathbb{G}$ is $\mathbb{Q}$. The “free” should be “torsion-free”.