Proof. For $\tau \in \mathcal{P}_0$ , it is easy to check that $(1-\tau ) (\Lambda ^*_\tau )=\Lambda _\tau $ . It follows that $\det (\Lambda ^{\tau })=\det (\Lambda _\tau )=\prod m^{a_m}$ , where $\prod m^{a_m}$ is the frame shape of $\tau $ . The result then follows by a direct calculation (cf. Table 1 in Section 6).

Notation 4.5. Let X be an even lattice and let $\beta \in {\mathbb Q} \otimes _{\mathbb Z} X$ . Denote

$$ \begin{align*}X_{\beta} = \{ x\in X \mid \langle x, \beta \rangle \in {\mathbb Z}\}.\end{align*} $$

Suppose that $\langle \beta ,\beta \rangle =2k/n$ for some positive integers k and n with $GCD(k,n)=1$ . Let $\bar {X}=\{x\in X\mid \langle x,n \beta \rangle \in {\mathbb Z}\}$ . For $x\in \bar {X}$ , we define $x_{[\beta ]} = x- m\beta $ if $\langle x, n\beta \rangle \equiv mk \mod n$ , $0\leq m\leq n-1$ and $\bar {X}_{[\beta ]} = \{ x_{[\beta ]}\mid x \in \bar {X}\}$ . Then $\bar {X}_{[\beta ]}$ is also an even lattice and $\det (\bar {X})=\det (\bar {X}_{[\beta ]})$ .

Proof. If $m\varphi (\sqrt {\ell }\beta )\in \Lambda $ , then $m\varphi (\sqrt {\ell }\beta )\in \Lambda ^{\tau }=\varphi (\sqrt {\ell }(\Lambda ^{\tau })^*)$ , which is equivalent to $m\sqrt {\ell }\beta \in \sqrt {\ell }(\Lambda ^{\tau })^{\ast }$ (or equivalently, $m\beta \in (\Lambda ^{\tau })^{\ast }$ ).

Since $[\Lambda ^\tau : \Lambda ^\tau _\beta ]=n ( = [ (\Lambda ^\tau _\beta )^* : (\Lambda ^\tau )^*]) $ and $(\Lambda ^\tau _\beta )^* = (\Lambda ^\tau )^* +{\mathbb Z} \beta $ , $m\beta \in (\Lambda ^{\tau })^{\ast }$ implies n divides m.

For the second statement, it suffices to show $n\sqrt {\ell } \varphi (\beta ) \in \Lambda ^\tau $ . Let $k=|\tau |$ . Then $\widehat {\tau }^k = \exp (2\pi i\delta (0))$ for some $\delta \in {\mathbb C} \Lambda ^\tau $ . Since $n=|\tau | N_0=kN_0$ and $\tilde {g}^n= \widehat {\tau }^n\exp (2\pi i\beta (0))^n=1$ , we have $n\beta -N_0 \delta \in \Lambda ^\tau $ .

As $\widehat {\tau }$ is a standard lift, $\widehat {\tau }(e^\gamma )=e^\gamma $ for any $\gamma \in \Lambda ^\tau $ . Thus, $\langle \delta , \Lambda ^\tau \rangle \subset {\mathbb Z}$ and $\delta \in (\Lambda ^\tau )^*$ . By using the isometry $\varphi $ , we have $n\sqrt {\ell } \varphi (\beta ) - N_0 \sqrt {\ell } \varphi (\delta ) \in \sqrt {\ell } \varphi (\Lambda ^\tau ) < \sqrt {\ell } \varphi ((\Lambda ^\tau )^*)= \Lambda ^\tau $ . Since $\delta \in (\Lambda ^\tau )^*$ , we have $\sqrt {\ell } \varphi (\delta ) \in \Lambda ^\tau $ and $n\sqrt {\ell } \varphi (\beta ) \in \Lambda ^\tau $ as desired.

Proof. Since $\tau $ fixes $\tilde {\beta }=\varphi (\sqrt {\ell }\beta )$ , we may assume that $\tau $ acts on $N=\Lambda ^{[\tilde {\beta }]}$ and $N^\tau = \Lambda ^\tau _{\tilde {\beta }} +{\mathbb Z} \tilde {\beta }$ . Note also that $\langle \beta , \beta \rangle \in (2/\ell ) {\mathbb Z}$ .

By Lemma 4.6, we have $L= \Lambda _\beta ^\tau +{\mathbb Z} \ell \beta $ . Thus,

$$ \begin{align*} L^* & = \{ x\in (\Lambda_\beta^\tau)^*\mid \langle x, \ell \beta \rangle \in {\mathbb Z}\}\\ &= \{ x\in (\Lambda^\tau)^*+{\mathbb Z} \beta \mid \langle x, \ell \beta \rangle \in {\mathbb Z}\}\\ & = < \{ x\in (\Lambda^\tau)^* \mid \langle x, \ell \beta \rangle \in {\mathbb Z}\}, \beta>. \end{align*} $$

Then

$$ \begin{align*} \sqrt{\ell} \varphi(L^*) & = < \{ \sqrt{\ell} \varphi(x) \in \sqrt{\ell} \varphi(\Lambda^\tau)^*\mid \langle \sqrt{\ell} \varphi(x) , \sqrt{\ell} \varphi(\beta) \rangle \in {\mathbb Z}\}, \sqrt{\ell} \varphi( \beta)>\\ &=< \{ y\in \Lambda^\tau \mid \langle y, \tilde{\beta} \rangle \in {\mathbb Z}\} , \tilde{\beta}>\\ &= \Lambda^\tau_{\tilde{\beta}} +{\mathbb Z} \tilde{\beta} = N^\tau \end{align*} $$

as desired.

Proof. Since $\tau $ fixes $\tilde {\beta }=\varphi (\sqrt {\ell }\beta )$ , we may assume that $\tau $ acts on $N=\Lambda ^{[\tilde {\beta }]}$ . Then $\tau $ induces an isometry $\tau '$ of $N=\Lambda ^{[\tilde {\beta }]}$ .

Suppose $N \cong \Lambda $ . Since all elements in $O(\Lambda )$ are determined by its frame shape up to conjugate, we may assume $\tau '$ is conjugate to $\tau $ . In particular, $N^{\tau '}$ is isometric to $\Lambda ^{\tau }$ and so $\sqrt {\ell }(N^{\tau '})^{\ast } \cong N^{\tau '}$ . By Theorem 4.9, $N^{\tau '}\cong \sqrt {\ell }L^{\ast }$ and so $(N^{\tau })^{\ast }= \frac {1}{\sqrt {\ell }}L$ . Then $\Lambda ^{\tau } \cong \sqrt {\ell }(\Lambda ^{\tau })^{\ast }\cong \sqrt {\ell }(N^{\tau '})^{\ast }\cong L$ . It is not possible since $\det (L)=\det (\Lambda ^{\tau }). |\tau |^2$ .

Proof. Let $ e^u \otimes t \in \mathcal{G}_{j,k_j}$ be a short root vector. Then $\langle u,u\rangle =\frac {2}{r_jk_j}$ and we have

$$ \begin{align*}\langle u,u\rangle=\frac{2}{r_jk_j}=\frac{2}{r_jh_j^{\vee}} \times \frac{h_j^{\vee}}{k_j} =\frac{2}{r_jh_j^{\vee}}\times \frac{N_0}{K_0-N_0}.\end{align*} $$

Since $\langle u,u\rangle \in \frac {{\mathbb Z}}{\ell }$ and $\mathrm {LCM}(\{r_jh_j^{\vee }/N_0:j\})=|\tau |$ , we have that $K_0-N_0$ divides $\ell /|\tau |$ .

Proof. Let u be a simple short root of $\mathcal{G}_j$ . Since $r_jh^{\vee }_j=n=|\tau |N_0$ , we have $\langle u, u\rangle = \frac {2}{|\tau |(K_0-N_0)}$ and $e^u \otimes t \in T^1_1$ . It implies $\langle u,u\rangle /2 \in 1/\ell +{\mathbb Z}$ . Note also that $\ell / |\tau |= 1$ or $2$ and $ |\tau |(K_0-N_0)$ divides $\ell $ . Therefore, $\ell = |\tau |(K_0-N_0)$ and $\langle u,u\rangle =2/\ell $ .

Proof. It follows from the fact that $\Lambda _{\tau } \cong L_B(C)$ and $\mathcal {D}(\Lambda _{\tau })$ has only one nonzero singular element of order $|\tau |$ , up to isometry if $\tau \in \mathcal{P}_0\setminus \{2C\}$ (see Appendix A).

Proof. For $\tau \in \mathcal{P}_0$ , it is easy to check that the frame shape of $\tau $ contains a positive power of $|\tau |$ . Therefore, $N_{\tau }$ contains a root system $A_{|\tau |-1}$ whose Coxeter number is $|\tau |$ .

Proof. For $\lambda \in \Lambda $ , let $\lambda ^* = (\lambda , 1, \frac {\langle \lambda ,\lambda \rangle }{2}-1)\in \Lambda +\Pi _{1,1}$ , which is called a Leech root. Set $\hat {\beta }= (\beta ,1,\frac {\langle \beta ,\beta \rangle }{2})\in {\mathbb Q} (\Lambda + \Pi _{1,1})$ . Note that $\hat {\beta }$ is an isotropic vector. Then $\lambda ^*-\hat {\beta } = (\lambda -\beta , 0 , \frac {\langle \lambda ,\lambda \rangle }{2}-1-\frac {\langle \beta ,\beta \rangle }{2})$ and we have

$$ \begin{align*} \langle \lambda-\beta,\lambda-\beta\rangle =\langle \lambda^*-\hat{\beta}, \lambda^*-\hat{\beta}\rangle =\langle \lambda^*, \lambda^* \rangle + \langle \hat{\beta},\hat{\beta}\rangle - 2\langle \lambda^*, \hat{\beta} \rangle =2-2\langle \lambda^*, \hat{\beta} \rangle. \end{align*} $$

Thus, $\langle \lambda -\beta ,\lambda -\beta \rangle =2$ if and only if $\langle \lambda ^*, \hat {\beta } \rangle =0$ .

Suppose that $\beta $ is not a deep hole. Then there is a $\delta \in \Lambda $ such that $\langle \delta -\beta ,\delta -\beta \rangle <2$ . In this case, $\langle \delta ^*, \hat {\beta } \rangle>0$ . Since $\delta ^{\ast }\not \in \{\lambda ^{\ast }\mid \lambda \in S\}$ and the minimal squared norm of $\Lambda $ is $4$ , $\langle \delta ^{\ast },\lambda ^{\ast }\rangle \leq 0$ for all $\lambda \in S$ . However, since $\tilde {S}$ contains an affine fundamental root system, there exists a subset $S'\subset S$ and positive real numbers $n_{\lambda }\in {\mathbb R}_{\geq 0}$ such that $\sum _{\lambda \in S'}n_{\lambda }\lambda = m \beta $ , where $m=\sum _{\lambda \in S'} n_{\lambda }$ . By direct calculations, we have

$$ \begin{align*} \sum_{\lambda\in S'}n_{\lambda}\left( \frac{\langle\lambda, \lambda\rangle}2 -1\right) &= \sum_{\lambda\in S'}n_{\lambda} \frac{\langle\lambda, \lambda\rangle -\langle \lambda-\beta, \lambda -\beta\rangle}2\\ & = \sum_{\lambda\in S'}n_{\lambda} \left( \langle\lambda, \beta\rangle -\frac{\langle \beta,\beta\rangle}2\right) =\frac{m\langle \beta, \beta\rangle}2. \end{align*} $$

Thus, we also have

$$\begin{align*}\sum_{\lambda\in S'}n_{\lambda}\lambda^*= \left ( \sum_{\lambda\in S'}n_{\lambda}\lambda, \sum_{\lambda\in S'}n_{\lambda}, \sum_{\lambda\in S'}n_{\lambda}\left( \frac{\langle\lambda, \lambda\rangle}2 -1\right)\right)= m \hat{\beta}, \end{align*}$$

but it implies $\langle \delta ^{\ast }, \hat {\beta }\rangle \leq 0$ , which is a contradiction.

Proof. By Lemma 4.14, it suffices to show that there is j such that $r_jh_j^\vee =n$ .

Recall that $\mathrm {LCM}(r_jh^{\vee }_j \mid j)=n$ . If $|\tau |$ is a prime power or $V_1$ is simple, then there is j such that $r_jh_j^{\vee }=n$ . Therefore, we may assume that $|\tau |$ is not a prime power and $\mathrm {rank}(V_1)>4$ ; that is, $\tau $ is $6E=1^22^23^26^2$ or $6G=2^36^3$ and $\mathrm {rank}(V_1)=8$ or $6$ . We have already shown $N_0| h_j^\vee $ for all j. By Corollary 5.6, we may also assume $N_0\not =1$ . Suppose there is no j such that $r_jh_j^\vee =n=6N_0$ . Then there exist k and l such that $r_kh_k^{\vee }=3N_0$ and $r_lh_l^\vee =2N_0$ . If one of $\mathcal{G}_j$ is of type $G_2$ , then $N_0=4$ and there is also a component $\mathcal{G}_i$ such that $r_ih_i^{\vee }=8$ ; that is, $\mathcal{G}_i=A_7$ , or $C_3$ . Another possible component is $A_3$ . Therefore, possible choices for $\tau $ and $V_1$ are $\tau =6E$ and $V_1=G_2C_3^2$ , or $G_2C_3A_3$ . However, since $\dim V_1=120\not =14+2\times 21$ nor $14+21+15$ , we have a contradiction. Therefore, there is no component of type $G_2$ . In order to get $r_kh_k^{\vee }=3N_0$ , $r_k=1$ , $h_k^\vee =3N_0$ and $V_1$ contains a component of type $A_{3N_0-1}$ . Since $N_0\not =1$ , $\mathrm {rank}(V_1)\leq 8$ and $V_1$ has a component with $r_ih_i^\vee =2N_0$ , we have $N_0=2$ and $V_1=A_3+A_5$ (i.e., $\mathrm {rank}(V_1)=8$ and $\tau =6E$ ). However, $\dim V_1=72\not =15+35$ , which is a contradiction. This completes the proof of Proposition 5.8.

Notation 5.9. For an even lattice K, we use $R(K)$ to denote the sublattice generated by $K_2=\{a\in K\mid \langle a,a\rangle =2\}$ . A component lattice of $R(K)$ means a sublattice generated by an irreducible component of the roots in $K_2$ .

Proof. If $\tilde {X}\not =X$ and $\tau $ does not act on $\tilde {X}$ trivially, then there are $\mu _1\in X$ and $\mu _2\in \tilde {X}-\tilde {X}^{\tau }$ such that $\langle \mu _1,\mu _2\rangle =-1$ . Since $\mu _2\not \in N^{\tau }$ and $\mathrm {rank}(R(N_{\tau }))=\mathrm {rank}(N_{\tau })$ (see Remark 5.5), there is a $\mu _3\in R(N_{\tau })$ such that $\langle \mu _2,\mu _3\rangle =-1$ . Since $\mu _3\in \tilde {X}$ , $\mathrm {rank}(\tilde {X})\geq \mathrm {rank}(X)+2$ , as we desired.

Proof. Suppose $N_0$ is even. Then $K_0$ is odd since $\mathrm {GCD}(K_0,N_0)=1$ . Since $K_0-N_0=1$ or $2$ , $K_0=N_0+1$ . Since $|\tau |=2$ , there are no components of type $G_2$ . Moreover, $N_0$ is even and $n/N_0=2$ ; thus, $4|n$ and there are no full components of type B and F. Therefore, the only full components of $V_1$ are of type $C_{N_0-1,1}$ . Since $N_0$ divides the dual Coxeter numbers, the other components are of the type $A_{N_0-1}$ or $D_{N_0/2+1}$ . Recall that $\mathrm {rank}(\Lambda ^\tau )=12$ or $16$ and $N_0$ is even. By a direct calculation, it is easy to verify that the possible cases are only $V_1=C_{3,1}^a A_{3,1}^b$ ( $N_0=4$ ) with $a+b=4$ and $C_7D_5$ ( $N_0=8$ ). Since $\dim V_1=24(N_0+1)$ , none of them is possible. Note that $\dim C_3=21, \dim A_3=15, \dim C_7=105$ and $ \dim D_5=45$ .

Proof. Without loss of generality, we may assume that $\mathcal{G}_1$ is a full component.

Case 1: If $\mathcal{G}_1=F_4$ , then $\dim F_4=52$ and $n=18$ and so the possible choices for $(\tau , \mathrm {rank}\Lambda ^{\tau },\dim V_1)$ are $(2A,16,240), (2C,12,132)$ , $(6G,6,60), (6E,8,72)$ , and the possible choices for $(\mathcal{G}_j, \dim \mathcal{G}_j)$ are $(F_4,52)$ , $(C_8,136)$ , $(A_8, 80)$ and $(A_2,8)$ . Hence the possible cases are $\tau =2A$ and $V_1=F_4^2C_8$ ; $\tau =2C$ and $V_1=F_4A_8$ ; and $\tau =6G$ and $V_1=F_4 A_2$ .

From now on, we may assume that there are no full components of type $F_4$ .

Case 2: If $\mathcal{G}_1=G_2$ , then $n=12$ and $|\tau |=3 (N_0=4, \dim V_1=120)$ or $|\tau |=6$ ( $N_0=2, \dim V_1=72$ or $60$ ). If $|\tau |=3$ , then $\tau =3B$ , $4|h^\vee _i$ , $h_i^\vee < 12$ and $\mathrm {rank}(\Lambda ^\tau )=12$ . Then $V_1=G_2^a+A_3^b$ and $2a+3b=12, \dim V_1=14a+15b=120$ , which has no solution. If $|\tau |=6$ , then $V_1=G_2^a+C_5^b+A_1^c+A_3^d+A_5^e$ . Since $\dim (V_1)=60$ or $72$ and $\mathrm {rank}(V_1) (=2a+5b+c+3d+5e) =6$ or $8$ , the only solution is $V_1=C_5G_2A_1$ .

From now on, we may assume that all full components are of type $C_m$ or of type $B_m$ .

Case 3: If $\mathcal{G}_1=C_8$ , then $n=18$ and $|\tau |=2$ or $6$ .

(3.1) If $|\tau |=2$ , then $N_0=9$ , and so possible components are $C_8, B_5, A_8$ ; thus, $V_1=C_8^2$ or $C_8A_8$ but $\dim (V_1)=24K_0=240$ . Neither one is possible.

(3.2) If $|\tau |=6$ , then $N_0=3$ and $V_1=C_8$ , but $\dim V_1=96\not =\dim C_8$ , a contradiction.

Case 4: If $\mathcal{G}_1=C_{2m}$ and $0<m\not =4$ , then $n=2(2m+1)$ . First, we show that $m<7$ . Suppose $m\geq 7$ . Then $\mathrm {rank}(V_1)\geq \mathrm {rank}(C_{2m})\geq 14$ and so $\tau =2A$ and $N_0=2m+1$ and $m=7$ or $8$ . In this case, $\dim V_1=24(2m+2)$ and $\dim C_{2m}=2m(4m+1)$ . If $m=7$ , then there are no components $\mathcal{G}_j$ of rank 2 (or less than 2) with $h_j^\vee $ divisible by $N_0=15$ . So, $m=8$ and $V_1=C_{16}$ ; however, $\dim V_1=24(N_0+1)=24\times 18\not = \dim C_{16}=16\times 33$ .

For $m<7$ and $m\neq 4$ , we have $(2m+1,3)=1$ , $N_0=2m+1$ and $|\tau |=2$ . So, the possible components of $V_1$ are $C_{2m}, A_{2m}, B_{m+1}$ . Since $K_0-N_0=1$ for $\tau =2A$ and $K_0-N_0=2$ for $\tau =2C$ , $\dim V_1=24(2m+2)$ for $\tau =2A$ and $\dim V_1=12(2m+3)$ for $\tau =2C$ . If $\tau =2A$ , then by solving the relation $2ma+2mb+(m+1)c=16$ and $2m(4m+1)a+2m(2m+2)b+(m+1)(2m+3)c=24(2m+2)$ and $a>0$ and $b,c\geq 0$ , we have $V_1=C_{10}B_6, C_6^2B_4, C_4^4$ .

If $\tau =2C$ , then by solving the relation $2ma+2mb+(m+1)c=12$ and $2m(4m+1)a+2m(2m+2)b+(m+1)2m+3)=12(2m+3)$ , we have $V_1=C_{4}A_{4}^2$ or $C_{2}^6$ .

We next assume $\mathcal{G}_1=C_{2m+1}$ . Then since $2m+2$ is even, $|\tau |\not =2$ .

Case 5: If $\mathcal{G}_1=C_3$ , then $n=8$ and $|\tau |=4$ and $N_0=2$ and $\mathcal{G}$ is a direct sum of $C_3$ or $A_1$ , or $A_3$ . Since $\dim C_3=21$ , $\dim A_3=15$ , $\dim A_1=3$ and $\mathrm {rank}(V_1)=10$ , it is impossible to have $\dim V_1=72$ , and we have a contradiction.

Case 6: If $\mathcal{G}_1=C_5$ , then $n=12$ and $(\tau ,N_0)=(4F,3),(6,2)$ . If $\tau =4F$ , then the other possible components are $A_2, D_4, A_5, C_2$ since there is no component of type $G_2$ . Since $\mathrm {rank}(V_1)=10$ is even, it has another component of type $C_5$ or $A_5$ , but since $\dim V_1=96$ , $\dim C_5=55$ , and $\dim A_5=35$ , we have a contradiction. If $|\tau |=6$ , then $\mathrm {rank}(V_1)=8$ or $6$ and possible components are $A_1, A_3$ . Since $\dim V_1=72$ or $60$ and $\dim A_1=3$ , $\dim A_3=15$ , we have a contradiction.

Case 7: If $\mathcal{G}_1=C_7$ , then $n=16$ and $(\tau ,N_0)=(4F,4),(8,2)$ . If $N_0=2$ , then $\dim V_1=72$ and $\dim C_7=105$ , a contradiction. If $\tau =4F$ , then $N_0=4$ , and the only solution is $V_1=C_7A_3$ .

Case 8: If $\mathcal{G}_1=C_{2m+1}$ and $m\geq 4$ , then $\mathrm {rank}(V_1)\geq 2m+1\geq 9$ and so $\mathcal{G}_1=C_9$ . However, since $\mathrm {rank}(V_1)=10$ and $N_0=5$ , we have a contradiction.

Case 9: We may assume that all full components are of type B. Say, $\mathcal{G}_1=B_m$ , then $n=2(2m-1)$ . Since $2m-1$ is odd, $|\tau |=2, 6$ .

If $|\tau |=6$ , then $N_0=1/3(2m-1)$ and $m\leq 8$ . Since $N_0\neq 1$ is odd, the only possible values for $(N_0,m)$ are $(3,5)$ and $(5,8)$ . If $\tau =6G$ , then $\mathrm {rank} V_1=6$ and $(N_0, m)=(3,5)$ . The only possible choice for $V_1$ is $B_5+A_1$ , but since the Coxeter number of $A_1$ does not divide $3$ , we have a contradiction. If $\tau =6E$ , then $\mathrm {rank} V_1=8$ , and the possible choice of $V_1$ is $B_8$ . Since $2C\not \in <6E>$ , $K_0=N_0+1=4$ and $\dim V_1=24K_0=96\not =\dim B_8=8\times 17$ , which is a contradiction.

Therefore, we have $|\tau |=2$ and $N_0=2m-1$ . Since $2m-1$ is odd, the non-full components are all of type $A_{2m-2}$ and so $V_1=B_m^{\otimes a}\oplus A_{2m-2}^{\otimes b}$ . We note $\dim B_m=m(2m+1)$ and $\dim A_{2m-2}=4m(m-1)$ . If $K_0=2m$ , then $am+b(2m-2)=16$ and $\dim V_1=48m=am(2m+1)+4bm(m-1)$ . It is easy to see that there is no solution for $a\in {\mathbb Z}_{> 0}$ and $b\in {\mathbb Z}_{\geq 0}$ . Hence, $K_0-N_0=2$ and $\tau =2C$ . In this case, $am+b(2m-2)=12$ and $\dim V_1=12(2m+1)=am(2m+1)+4bm(m-1)$ . The solutions are $b=0$ and $am=12$ ; therefore, $V_1=B_2^6, B_3^4, B_4^3, B_6^2$ , or $B_{12}$ , as we desired.

This completes the proof of Lemma 5.12.