- Page 6, second displayed equation in Remark 1.1: $\varepsilon_i(x)$ should be $\varepsilon_i(y)$ (twice).
- Page 11, Example 2.10: In line -2 spin($2r+1$) should be spin($2r$).
- Page 11, Example 2.11: "...if it is in the adjoint lattice $\mathbb{Z}^r$, and we will call it a..." should be replaced by "...if it is in the lattice $\mathbb{Z}^r$, and we will call it a...".
- Page 15, Definition 2.20: It is not necessary to assume that $\mathcal{C}$ is of finite type.
- Page 18, line 5 of the proof of Proposition 2.29. The equation $e_i(w\otimes z)=w\otimes e_i(z)=x\otimes y$ should read $e_i(z\otimes w)=z\otimes e_i(w)=x\otimes y$.
- Page 20: For the definition of a crystal isomorphism it also needs to be assumed that the inverse of $\psi$ is a crystal morphism.
- Page 22, Section 2.4: It is assumed in this section that the crystals are seminormal.
- Page 24, Proposition 2.37: The crystal in this proposition is assumed to be seminormal.
- Page 25: For the definition of an isogeny, we assume that (in addition to the stated conditions) $\langle m(\lambda),m(\alpha)\rangle = \langle\lambda,\alpha\rangle$ when $\lambda$ is in $\Lambda$ and $\alpha$ is in $\Phi$.
- Page 27: First bullet point should read: If $\alpha_j$ is the longer root, then $|\alpha_j|=\sqrt2|\alpha_i|$. Second bullet point should read: If $\alpha_j$ is the longer root, then $|\alpha_j|=\sqrt3|\alpha_i|$.
- Page 29: Exercise 2.6(ii) it is assumed that the crystal has a unique highest weight element.
- Page 36: (5 lines before exercises): $v\mapsto R(T)$ should be $T\mapsto \hbox{RR}(T)$.
- Page 38, Lemma 4.1: The assumption that $\mathcal{B}$ is upper seminormal is needed.
- Page 40, Example 4.3, Row 4: solid arrows correspond to $f_2$ and dashed arrows correspond to $f_1$
- Page 40, Example 4.3, Row 8: $e_2e_1x=e_2e_1x$ should be replaced by $e_1e_2x=e_2e_1x$
- Page 41, Lemma 4.4: Add the assumption that $\mathcal{C}$ is of finite type.
- Page 45, Propositions 4.7 and 4.9: Assume that the crystal is a Stembridge crystal (rather than a weak Stembridge crystal).
- Page 46, middle: $I \otimes A\cong I$ should be $\cong A$.
- Page 47, line 1: $w=e_i(y)$ should be replaced by $w=e_j(y)$.
- Page 47, Case 1, second display: $\varepsilon_j(e_iy)=\varepsilon_i(y)$ should be replaced by $\varepsilon_j(e_iy)=\varepsilon_j(y)$.
- Page 48, Case 1A: In the diagram replaced $e_ix\otimes e_j^2e_ix$ by $e_ix\otimes e_j^2e_iy$.
- Page 48, Case 1A: In the last line beneath the diagram replace $e_j^2e_ix\neq 0$ by $e_j^2e_iy\neq 0$.
- Page 49, Case 1B: In the diagram, $e_jx\otimes e_i^2e_jx$ should be $e_jx\otimes e_i^2e_jy$.
- Page 49, Case 1B: In the diagram, $e_ix\otimes y$ should be $x\otimes e_i$.
- Page 52, middle: "...then $\text{wt}(u)$ is a linear combination with nonnegative integer coefficients of the simple roots" should be "...then $\text{wt}(u)-\text{wt}(x)$ is a linear combination with nonnegative integer coefficients of the simple roots".
- Page 53, proof of Theorem 4.13: "and since $\Omega$ is closed" to be replaced by "and since $S$ is closed"
- Page 62, Proposition 5.7: The assumption that $\mathcal{V}$ is connected is not needed.
- Page 63, Theorem 5.9 and Theorem 5.10: Both theorems require the assumption that the ambient crystal is bounded above.
- Page 79, definition of crystal involution: $e_i(Sx) = f_{i'}(x)$ should be $S(e_i(Sx)) = f_{i'}(x)$ and similarly the next equality.
- Page 84, Condition (2) before Example 6.5 needs to hold for all $1\leqslant j< r$.
- Page 98, description of Schensted insertion: "As before, $k$ is bumped by $i$" is replaced by "As before, $k$ is bumped by $j$".
- Page 100: "since it is a special case of Theorem 7.14" is replaced by "since it is a special case of Theorem 7.13"
- Page 105, line minus 3: Otherwise, it bumps the
**smallest**entry that is $\geqslant j$. - Page 172, Exercise 12.6: "Prove that $\mathcal{B}_1\times\mathcal{B}_2\times\mathcal{B}_1$ is not connected" should be replace by "Prove that $\mathcal{B}_1\times\mathcal{B}_2 \times\mathcal{B}_1$ is connected."
- Page 185, paragraph beginning "We consider the first case $\varepsilon_i(x')=\varepsilon_i(x)=0$ now." This argument has a gap but can be fixed as follows.

Note that this implies that $z=x$ and $z'=x'$, so that $z'=x'=e_jx = e_jz$. We also have $\varepsilon_i(x)=\varepsilon_i(\psi_i(x))$, which equals the maximum of $\varepsilon_i(y)$ and $t+\varepsilon_i(y)-\varphi_i(y)$. In particular $\varepsilon_i (y) = 0$ and by similar arguments $\varepsilon_i(y')=0$. Since $\varepsilon_i^\star(x)=t=t'=\varepsilon_i^\star(x')$ and $\varepsilon_i (x') = \varepsilon_i (x) = 0$, (14.8) implies (14.7) provided we can show $\varphi_i(y') - \varphi_i(y) = \varphi_i^\star(z') - \varphi_i^\star(z)$. But $y'=e_j y$, $z' = e_jz$ and in addition $\varepsilon_i(y)=\varepsilon_i(y')$ and $\varepsilon_i^\star(z) = \varepsilon_i^\star(x)=\varepsilon_i^\star(x') = \varepsilon_i^\star(z')$, so both $\varphi_i(y') - \varphi_i(y)$ and $\varphi_i^\star(z') - \varphi_i^\star(z)$ equal $\langle\alpha_j,\alpha_i^\vee\rangle$.

- Page 199, top figure: the entries in the right column should be $(a-1,b-1,a)$, $(a,b-1,a)$ and $(a+1,b-1,a)$.