FC11 — Peters–Sterk, Symmetric and Quadratic Forms (June 2024)
FC11 — Peters & Sterk, Symmetric and Quadratic Forms.
This chapter contains 5 definitions, 1 theorem, 5 propositions, 2 lemmas, 10 examples and 1 construction.
Nobody has checked this chapter against Lean yet, so the statements below say nothing about whether they are formalized.
Definition 11.1.1
Definition 11.1.1. Let \(p\) be an odd prime. • A homogeneous symmetric \(p\) -adic lattice of rank \(r \geq 1\) and exponent \(k\) is in homogeneous normal form if it is of the form \(L _ { r , u } ( p ^ { k } )\) , where \[ L _ { r , u } = \langle u \rangle \oplus \oplus ^ { r - 1 } \langle 1 \rangle , \quad u \in \mathbb { Z } _ { p } ^ { \times } , \] a unimodular lattice. • A symmetric \(p\) -adic lattice is in normal form if it is an orthogonal direct sum of homogeneous normal forms. • A su \(p\) -primary symmetric torsion form is in normal form if it is an orthogonal of forms \(b _ { r , u } ^ { \# } ( p ^ { - k } )\) , the symmetric discriminant form of \(L _ { r , u } ( p ^ { k } )\) , \(k \geq 1\)
Lemma 11.1.2
Lemma 11.1.2. 1. For fixed \(r\) there are exactly two isometry classes for \(L _ { r , u }\) distinguished by \(u\) being a square or a non-square modulo \(p\) . 2. Every non-degenerate \(p\) -adic lattice admits a normal form of the above shape.
Definition / construction
For a \(p\)-primary torsion form in the normal form of Proposition 9.4.1, associate the \(p\)-adic lattice \(L_{b^\#}=\bigoplus_i L_{r_i,u_i}(p^{k_i})\).
Proposition 11.1.3
Proposition 11.1.3. Let \(p\) be an odd prime and let \(( G , b ^ { \# } )\) be a non-degenerate \(p\) -primary torsion symmetric group of length \(\ell ( G )\) with \(b ^ { \# }\) in normal form (11.1). Then: 1. The lattice \(L _ { b ^ { \# } }\) given by (11.2) is up to isometry the unique non-degenerate \(p\) -adic lattice of rank \(\ell ( G )\) and discriminant form \(b ^ { \# }\) . Its discriminant equals \(\begin{array} { r } { | G | \cdot \prod _ { k = 1 } ^ { s } u _ { k } = | G | \cdot \delta ( b ^ { \# } ) } \end{array}\) , where \(\delta ( b ^ { \# } )\) is the reduced discriminant of \(b ^ { \# }\) . 2. The discriminant form map \[ \left\{ \begin{array} { c c } { { i s o m . ~ c l a s s e s ~ o f ~ p / - a d i c } } \\ { { s y m m e t r i c ~ l a t t i c e s ~ o f ~ r a n k ~ r } } \end{array} \right\} \longrightarrow \left\{ \begin{array} { c c } { { i s o m . ~ c l a s s e s ~ o f ~ p / - p r i m a r y ~ t o r s i o n } } \\ { { s y m m e t r i c ~ f o r m s ~ o f ~ l e n g t h ~ r } } \end{array} \right\} \] is injective (and hence bijective).
Definition / construction
For a non-degenerate symmetric \(p\)-primary torsion form \(b^\#\), \(L_{b^\#}\) denotes the uniquely determined rank-\(\ell(G)\) \(p\)-adic lattice in the associated normal-form isometry class; this notation is used even when \(b^\#\) itself is not displayed in normal form.
Proposition 11.1.4
Proposition 11.1.4 (Normal forms for non-degenerate \(p\) -adic lattices, \(p\) odd). Let \(p\) be an odd prime, \(( G , b ^ { \# } )\) a non-degenerate \(p\) -primary symmetric torsion form, \(r\) an integer \(\geq \ell ( G )\) and \(t = r - \ell ( G )\) . 1. Let ?? be a non-square modulo \(p\) . A non-degenerate \(p\) -adic lattice \(L\) of rank \(r\) and with discriminant form \(( G , b ^ { \# } )\) is isometric to one of the following two lattices: \[ \begin{array}{c} \begin{array} { r } { \left\{ L \simeq L _ { b ^ { \# } } \oplus \mathbb { O } ^ { t } \langle 1 \rangle \qquad i f \operatorname { d i s c } ( L ) = \operatorname { d i s c } ( L _ { b ^ { \# } } ) = | G | \cdot \delta ( b ^ { \# } ) \qquad \right.} \\ { L \simeq L _ { b ^ { \# } } \oplus \langle \epsilon \rangle \oplus ^ { t - 1 } \langle 1 \rangle \quad i f \operatorname { d i s c } ( L ) = \epsilon \cdot \operatorname { d i s c } ( L _ { b ^ { \# } } ) = | G | \cdot \epsilon \cdot \delta ( b ^ { \# } ) . } \end{array} \end{array} \] 2. Let \(\widetilde { L }\) be a non-degenerate integral lattice of rank \(r\) . Then the discriminant form of \(\widetilde { L }\) determines the localization \(\widetilde { L } _ { p }\) up to isometry.
Example / comparison
We take \(p = 7\) . The squares modulo 7 are \(1 , 2 , 4\) and the nonsquares are \(3 , 5 , 6\) . Consider the torsion form \(b ^ { \# } = \langle 3 \cdot 7 ^ { - 1 } \rangle \oplus \langle 7 ^ { - 1 } \rangle \oplus \langle 5 \cdot 7 ^ { - 2 } \rangle\) . This is already in normal form and it is the discriminant form of \(L _ { b \# } = \langle 3 \cdot 7 \rangle \textcircled { \scriptsize { \oplus } } \langle 7 \rangle \textcircled { \scriptsize { \oplus } } \langle 5 \cdot 7 ^ { 2 } \rangle\) . Its discriminant equals \(7 ^ { 4 }\) (up to squares of units) and so, for example, the normal form of a 7-adic lattice \(L\) of rank 8 with discriminant \(5 \cdot 7 ^ { 4 }\) and discriminant symmetric form \(b ^ { \# }\) is \(L = \left. 3 \cdot 7 \right. \mathbb { \odot } \left. 7 \right. \mathbb { \odot } \left. 5 \cdot 7 ^ { 2 } \right. \mathbb { \odot } \left. 5 \right. \mathbb { \odot } \mathbb { \oplus } ^ { 4 } \langle 1 \rangle\) .
Example / comparison
Let us consider the normal forms of the \(p\) -adic localizations of the A-D-E root lattices, where \(p\) is an odd prime. First of all we consider the unimodular root lattice \(E _ { 8 }\) . Since \(\mathrm { d i s c } ( E _ { 8 } ) = 1\) , all of its localizations have discriminant 1. Hence, the normal form of \(E _ { 8 } \otimes \mathbb { Z } _ { p }\) is the diagonal \(p\) -adic lattice \(\mathbb { O } ^ { 8 } \langle 1 \rangle\) . The normal forms of the other root lattices depend on the odd prime \(p\) . To determine these, one may use Table 4.1.1. As an example, consider \(A _ { 5 }\) , which has discriminant 6 and discriminant group isomorphic to \(\mathbb { Z } / 6 \mathbb { Z } = \mathbb { Z } / 2 \mathbb { Z } \times \mathbb { Z } / 3 \mathbb { Z }\) . If \(p \neq 2 , 3\) the form \(A _ { 5 } \otimes \mathbb { Z } _ { p }\) is unimodular and its discriminant is a unit in \(\mathbb { Z } _ { p }\) which may or may not be a square modulo \(p\) . For instance, 6 is a non-square modulo 7 and so \(A _ { 5 } \otimes \mathbb { Z } _ { 7 } \cong \langle 6 \rangle \oplus \oplus ^ { 4 } \langle 1 \rangle\) , while \(A _ { 5 } \otimes \mathbb { Z } _ { 5 } \cong \oplus ^ { 5 } \langle 1 \rangle\) since 6 is a square modulo 5. For \(p = 3\) we need the 3-adic discriminant symmetric form of \(A _ { 5 }\) which is the restriction of the discriminant form of \(A _ { 5 }\) to the 3-primary part by (1.20). The discriminant form of \(A _ { 5 }\) is \(\langle 5 / 6 \rangle\) . The 3-primary part of \(\mathbb { Z } / 6 \mathbb { Z }\) is isomorphic to \(\mathbb { Z } / 3 \mathbb { Z }\) and generated by the class of 2 in \(\mathbb { Z } / 6 \mathbb { Z }\) , so that the 3-adic discriminant symmetric form \(b ^ { \# }\) is determined by its value on the pair \(( 2 , 2 )\) , which is \(2 ^ { 2 } \cdot 5 / 6 \equiv 1 / 3\) modulo \(\mathbb { Z }\) . So \(b ^ { \# } = \langle 1 / 3 \rangle\) and \(L _ { b \# } = \langle 3 \rangle\) . Hence \(A _ { 5 } \otimes \mathbb { Z } _ { 3 }\) has normal form \(\langle 3 \rangle \oplus \langle 2 \rangle \oplus \oplus ^ { 3 } \langle 1 \rangle\) since 2 is a non-square in \(\mathbb { Z } _ { 3 }\) .
Lemma 11.2.1
Lemma 11.2.1 (Relations between dyadic lattices and 2-primary torsion forms). Let \(u , u ^ { \prime } , u ^ { \prime \prime }\) be units in \(\mathbb { Z } _ { 2 }\) . Then \(v = u u ^ { \prime } + u u ^ { \prime \prime } + u ^ { \prime } u ^ { \prime \prime }\) is \(a\) 2-adic unit with \(v \equiv 3 { \mathrm { ~ m o d ~ } } 8\) or \(v \equiv - 1\) mod 8 and the following relations hold between symmetric 2-adic lattices: \[ \begin{array} { r } { U _ { k } \oplus U _ { k } \simeq V _ { k } \oplus V _ { k } , \quad k \geq 0 , \qquad } \\ { ( \langle u \rangle \oplus \langle u ^ { \prime } \rangle \oplus \langle u ^ { \prime \prime } \rangle ) ( 2 ^ { k } ) \simeq \left\{ \begin{array} { l l } { V _ { k } \oplus U _ { k } \oplus \langle ( u + u ^ { \prime } + u ^ { \prime \prime } ) \cdot 2 ^ { k } \rangle } \\ { \quad i n \ c a s e \ u u ^ { \prime } + u u ^ { \prime \prime } + u ^ { \prime } u ^ { \prime \prime } \equiv 3 \bmod 8 } \\ { U _ { k } \oplus \langle ( u + u ^ { \prime } + u ^ { \prime \prime } ) \cdot 2 ^ { k } \rangle } \\ { \quad i n \ c a s e \ u u ^ { \prime } + u u ^ { \prime \prime } + u ^ { \prime } u ^ { \prime \prime } \equiv - 1 \bmod 8 } \\ { \quad ( \langle u \rangle \oplus \langle u ^ { \prime } \rangle ) ( 2 ^ { k } ) \simeq ( \langle - 3 u \rangle \oplus \langle - 3 u ^ { \prime } \rangle ) ( 2 ^ { k } ) , \quad k \geq 0 } \end{array} \right. \quad k \geq 0 } \end{array} \] There are similar relations (I) – (III) between the corresponding discriminant forms.
Definition 11.2.2
Definition 11.2.2. • A homogeneous symmetric dyadic lattice of rank \(r\) and exponent \(k\) is in normal form if it is of the form \(L ( 2 ^ { k } )\) with \[ L = \mathbb { O } _ { i \leq a } \langle u ^ { ( i ) } \rangle \oplus \mathbb { O } ^ { b } V \mathbb { O } ^ { c } U , \quad a \leq 2 , b \leq 1 , c = \frac { 1 } { 2 } ( r - a ) - b \geq 0 , u ^ { ( i ) } \in \mathbb { Z } _ { 2 } ^ { \times } . \] For \(k = 0\) and \(L\) an even dyadic lattice, the cyclic forms are not present, that is, \(a = 0\) in this case. • A symmetric dyadic lattice is in normal form if it is the orthogonal direct sum of homogeneous normal forms. • A 2-primary symmetric (quadratic) torsion form is in normal form if it is the discriminant form of a symmetric (quadratic) dyadic lattice in normal form.
Proposition 11.2.3
Proposition 11.2.3. • Let \(( L , b )\) be a non-degenerate symmetric (quadratic) dyadic lattice. Then \(L\) is isometric to a normal form and so is its symmetric (quadratic) discriminant form. • Every 2-primary symmetric (quadratic) torsion form is isometric to a normal form. Moreover, there is a one-to-one correspondence between normal forms for symmetric and quadratic torsion forms on the same torsion group (with the same notation, as illustrated in the table below).
Construction / comparison
For dyadic discriminant forms, the symmetric and quadratic classifications do not coincide: distinct quadratic torsion forms can have the same polar form, and a fixed symmetric discriminant form of length at most \(2\) can come from several non-isometric rank-minimal dyadic lattices. In particular \(U_1\) and \(V_1\) have the same symmetric but different quadratic discriminant forms; for exponent \(k\ge2\) this discrepancy disappears.
Proposition 11.2.4
Proposition 11.2.4. Let \(( G , q ^ { \# } )\) be a quadratic 2-primary torsion group of length \(r\) decomposed into homogeneous normal form. Then the following two types of forms occur, each with different behaviour with respect to normal forms: Type 1. No cyclic orthogonal summands of order 2 split off from \(q ^ { \# }\) . Then there is a unique isometry class of a dyadic lattice \(L\) of rank \(r\) with discriminant form \(q ^ { \# }\) represented by the normal form \(L _ { q ^ { \# } }\) with discriminant \(\delta ( q ^ { \# } ) \cdot | G |\) . Type 2. Otherwise there are precisely two isometry classes ?? (??) , \(i = 1 , 2\) , of rank ?? lattices with discriminant form \(q ^ { \# }\) in normal form characterized by their discriminants, \(\mathrm { d i s c } ( L _ { q ^ { \# } } ^ { ( 1 ) } ) = \delta ( q ^ { \# } ) \cdot | G |\) , respectively \(\mathrm { d i s c } ( L _ { q ^ { \# } } ^ { ( 2 ) } ) = - 3 \cdot \delta ( q ^ { \# } ) \cdot | G |\) . Moreover the discriminant form map \[ \left\{ \begin{array} { c c } { { i s o m . \ c l a s s e s \ o f \ d y a d i c } } \\ { { q u a d r a t i c \ l a t t i c e s \ o f \ r a n k \ r } } \end{array} \right\} \longrightarrow \left\{ \begin{array} { c c } { { i s o m . \ c l a s s e s \ o f \ 2 - p r i m a r y \ t o r s i o n } } \\ { { q u a d r a t i c \ f o r m s \ o f \ l e n g t h \ r } } \end{array} \right\} \] is surjective, and injective on the preimage of type 1 forms, and two-to-one onto on the preimage of type 2 forms.
Example / comparison
From Table 11.2.2 we see that there are two normal forms of non-degenerate quadratic torsion forms on \(\oplus ^ { 5 } \mathbb { Z } / 2 ^ { 3 } \mathbb { Z }\) . First of all, \(\left. u \cdot 2 ^ { - 3 } \right. \textcircled { \pm } ^ { 2 } u _ { 3 }\) , the discriminant form of \(\langle u \cdot 2 ^ { 3 } \rangle \oplus ^ { 2 } U _ { 3 }\) , and, secondly, \(\left. u \cdot 2 ^ { - 3 } \right. \textcircled { \pm } u _ { 3 } \textcircled { \pm } v _ { 3 }\) , the discriminant form of \(\left. u \cdot 2 ^ { 3 } \right. \ @ U _ { 3 } \oplus V _ { 3 }\) .
Example / comparison
On \(\mathbb { Q } ^ { 6 } \mathbb { Z } / 2 \mathbb { Z }\) we have the following possibilities for non-degenerate quadratic torsion forms: • \(\mathbb { O } ^ { 3 } u _ { 1 }\) , the discriminant form of \(\mathbb { O } ^ { 3 } U _ { 1 }\) ; • \(\mathbb { O } ^ { 2 } u _ { 1 } \oplus v _ { 1 }\) , the discriminant form of \(\mathbb { O } ^ { 2 } U _ { 1 } \oplus V _ { 1 }\) ; • \(\left. u \cdot 2 ^ { - 1 } \right. \mathbb { O } \left. u ^ { \prime } \cdot 2 ^ { - 1 } \right. \mathbb { O } ^ { 2 } u _ { 1 }\) , the discriminant form of \(\left. u \cdot 2 \right. \textcircled { 1 } \left. u ^ { \prime } \cdot 2 \right. \textcircled { 1 } ^ { 2 } U _ { 1 }\) or of \(\left. - 3 u \cdot 2 \right. \textcircled { 1 } \left. u ^ { \prime } \cdot 2 \right. \textcircled { 1 } ^ { 2 } U _ { 1 }\) ; • \(\langle u \cdot 2 ^ { - 1 } \rangle \mathbb { O } \langle u ^ { \prime } \cdot 2 ^ { - 1 } \rangle \mathbb { O } u _ { 1 } \mathbb { O } v _ { 1 }\) , the discriminant form of \(\left. u \cdot 2 \right. \mathbb { \oplus } \left. u ^ { \prime } \cdot 2 \right. \mathbb { \oplus } U _ { 1 } \mathbb { \oplus } V _ { 1 }\) or of \(\left. - 3 u \cdot 2 \right. \textcircled { 1 } \left. u ^ { \prime } \cdot 2 \right. \textcircled { 1 } U _ { 1 } \textcircled { 1 } V _ { 1 }\) . We next use the above classification to find normal forms for discriminant forms of several integral lattices related to root lattices.
Example / comparison
The unimodular lattice \(E _ { 8 }\) has discriminant 1. Likewise, \(E _ { 8 } \otimes \mathbb { Z } _ { 2 }\) is unimodular with discriminant 1. Hence (see Example 10.2.3) it is isometric to \(\mathbb { Q } ^ { 4 } U \otimes \mathbb { Z } _ { 2 }\) . To find the discriminant form for \(L = E _ { 8 } ( 2 ^ { k } )\) , arguing as in Example 1.6.8.2, we see that ??#??⊗ℤ \(b _ { L \otimes \mathbb { Z } _ { 2 } } ^ { \# } = b _ { L } ^ { \# } \simeq \oplus ^ { 4 } u _ { k }\) .
Example / comparison
The lattice \(E _ { 7 }\) has 2-primary discriminant quadratic form \(q ^ { \# } = \langle 3 \cdot 2 ^ { - 1 } \rangle\) with ??(1) \(L _ { q ^ { \# } } ^ { ( 1 ) } = \langle 3 \cdot 2 \rangle\) \(L _ { q ^ { \# } } ^ { ( 2 ) } = \langle - 1 \cdot 2 \rangle\) . Then . Turni \(E _ { 7 } \otimes \mathbb { Z } _ { 2 }\) normal form , one then find \(\langle - 1 \cdot 2 \rangle \oplus \oplus ^ { 3 } U\) \(\operatorname { d i s c } ( E _ { 7 } ) = 2 = ( - 2 ) \cdot ( - 1 ) ^ { 3 }\) \(E _ { 7 } ( 2 ^ { k } )\) \(q _ { E _ { 7 } ( 2 ^ { k } ) } ^ { \# } =\) \(\langle - 2 ^ { - k - 1 } \rangle \oplus \oplus ^ { 3 } u _ { k }\) . The next task is to describe standard forms for dyadic lattices \(L\) whose ranks are not necessarily equal to the length of their discriminant group \({ \mathsf { d g } } _ { L }\) . Observe that a unimodular even \(\mathbb { Z } _ { 2 }\) -lattice must have even rank as follows from the normal form decomposition. So the unimodular summands that need to be added to the forms ????# or ??(1) , ??(2) can only be of rank two. This explains why \(\operatorname { r a n k } ( L ) - \ell ( G )\) is an even number in the statement of the main result, the analog of Proposition 11.1.4.
Proposition 11.2.6
Proposition 11.2.6 (Normal forms for quadratic dyadic lattices). Let \(( G , q ^ { \# } )\) . be a 2-primary quadratic torsion group, ?? an integer \(\geq \ell ( G )\) and \(t = r - \ell ( G )\) . 1. A non-degenerate even 2-adic lattice \(L\) of rank ?? and with discriminant quadratic form \(( G , q ^ { \# } )\) is isometric to one of the following two types of lattices: (a) If \(q ^ { \# }\) has no cyclic orthogonal summands of order 2 in its normal form decomposition, then \[ L \simeq \left\{ \begin{array} { l l } { { \oplus ^ { \frac { 1 } { 2 } t } U \oplus L _ { q \# } } } & { { \mathrm { ~ } i f \mathrm { ~ d i s c } ( L ) = \mathrm { d i s c } ( L _ { q \# } ) \cdot ( - 1 ) ^ { \frac { 1 } { 2 } t } } } \\ { { \oplus ^ { \frac { 1 } { 2 } t - 1 } U \oplus V \oplus L _ { q \# } } } & { { \mathrm { ~ } i f \mathrm { ~ d i s c } ( L ) = \mathrm { d i s c } ( L _ { q \# } ) \cdot 3 \cdot ( - 1 ) ^ { \frac { 1 } { 2 } t - 1 } . } } \end{array} \right. \] (b) Otherwise, that is, if \(q ^ { \# } \simeq \langle u \cdot 2 ^ { - 1 } \rangle \oplus q ^ { \prime }\) for some unit \(u \in \mathbb { Z } _ { 2 }\) , then \[ { \cal L } \simeq \oplus ^ { \frac { 1 } { 2 } t } U \oplus { \cal L } _ { q ^ { \# } } ^ { ( i ) } , \quad i = 1 , 2 . \] Which of the two cases occurs can be determined as follows: If \(\delta ( q ^ { \# } ) \cdot | G | \equiv\) \(( - 1 ) ^ { \frac { 1 } { 2 } t } \operatorname { d i s c } ( L )\) mod 4, then \(i = 1\) and otherwise \(i = 2\) . 2. Let \(\widetilde { L }\) be a non-degenerate even integral lattice of rank \(r\) . Then the discriminant form of \(\widetilde { L }\) determines the localization \(\widetilde { L } _ { 2 }\) uniquely up to isometry.
Example / comparison
For the first example with \(G = \oplus ^ { 5 } \mathbb { Z } / 2 ^ { 3 } \mathbb { Z }\) , \(\ell ( G ) = 5\) , there are two possible quadratic forms \(q ^ { \# }\) : • \(q ^ { \# } = \langle u \cdot 2 ^ { - 3 } \rangle \mathbb { \oplus ^ { 2 } } u _ { 3 }\) with \(L _ { q \# } = \langle u \cdot 2 ^ { 3 } \rangle \textcircled { \# } ^ { 2 } U _ { 3 }\) . Then either \(L \simeq \langle u \mathrm { \cdot 2 ^ { 3 } } \rangle \oplus ^ { 2 } U _ { 3 } \oplus ^ { 2 } U\) with discriminant \(u \cdot 2 ^ { 1 5 }\) or \(L \simeq \langle u \cdot 2 ^ { 3 } \rangle \oplus ^ { 2 } U _ { 3 } \oplus U \oplus V\) with discriminant \(- 3 u \cdot 2 ^ { 1 5 }\) . So in this case \(\mathrm { d i s c } ( L ) \neq - u , 3 u\) . • \(q ^ { \# } = \langle u \cdot 2 ^ { - 3 } \rangle \mathbb { \oplus } u _ { 3 } \oplus v _ { 3 }\) with \(L _ { q \# } = \langle u \cdot 2 ^ { 3 } \rangle \textcircled { \# } U _ { 3 } \textcircled { \# } V _ { 3 }\) . Then either \(L \simeq \langle u \cdot 2 ^ { 3 } \rangle \textcircled { \pm }\) \(U _ { 3 } \oplus V _ { 3 } \oplus ^ { 2 } U\) with discriminant \(- 3 u \cdot 2 ^ { 1 5 }\) or \(L \simeq \langle u \cdot 2 ^ { 3 } \rangle \oplus U _ { 3 } \oplus V _ { 3 } \oplus U \oplus V\) with discriminant \(u \cdot 2 ^ { 1 5 }\) (up to squares of units) and here also \(\mathrm { d i s c } ( L ) \neq - u , 3 u\) . So no rank 9 lattices \(L\) with \({ \mathsf { d g } } _ { L } = G\) and \(\mathrm { d i s c } ( L ) = - u , 3 u\) exist.
Example / comparison
For the second example, \(G = \oplus ^ { 6 } \mathbb { Z } / 2 \mathbb { Z }\) , \(\ell ( G ) = 6\) and there are four possible \(q ^ { \# }\) . We search for rank 10 dyadic lattices \(L\) with discriminant \(- 3 \cdot 2 ^ { 6 }\) , starting with the rank 6 lattices \(L _ { q ^ { \# } }\) in the first two cases and both forms ??(??)??# , ?? = 1, 2, in the last two cases: • \(L _ { q ^ { \# } } = \oplus ^ { 3 } U _ { 1 }\) has discriminant \(- 2 ^ { 6 }\) . It is impossible to get the required discriminant \(- 3 \cdot 2 ^ { 6 }\) from this since \(U \odot U\) has discriminant 1. • \(L _ { q ^ { \# } } = \oplus ^ { 2 } U _ { 1 } \oplus V _ { 1 }\) has discriminant \(3 \cdot 2 ^ { 6 }\) which is not possible for the same reason. Here ??(1) \(L _ { q ^ { \# } } ^ { ( 1 ) } = \left. u \cdot 2 \right. \mathbb { O } \left. u ^ { \prime } \cdot 2 \right. \mathbb { O } ^ { 2 } U _ { 1 }\) and one has \(\mathrm { d i s c } ( L _ { q ^ { \# } } ^ { ( 1 ) } ) = u u ^ { \prime } \cdot 2 ^ { 6 }\) . However, \({ \cal L } = { \cal L } _ { q ^ { \# } } ^ { ( 1 ) } \oplus ^ { 2 } U\) has discriminant \(u u ^ { \prime } 2 ^ { 6 }\) which is never equal to \(- 3 \cdot 2 ^ { 6 }\) since \(u , u ^ { \prime } \in \{ 1 , 3 \}\) . If \(L _ { q ^ { \# } } ^ { ( 2 ) } = \langle - 3 u \cdot 2 \rangle \oplus \langle u ^ { \prime } \cdot 2 \rangle \oplus ^ { 2 } U _ { 1 }\) , the discriminant is \(- 3 u u ^ { \prime } \cdot 2 ^ { 6 }\) and in order that \({ \cal L } = { \cal L } _ { q ^ { \# } } ^ { ( 2 ) } \oplus ^ { 2 } U\) , we must have \(u u ^ { \prime } = 1\) , i.e, \(u = u ^ { \prime }\) . Here ??(1) \(L _ { q ^ { \# } } ^ { ( 1 ) } = \left. u \cdot 2 \right. \oplus \left. u ^ { \prime } \cdot 2 \right. \oplus U _ { 1 } \oplus V _ { 1 }\) or \(L _ { q ^ { \# } } ^ { ( 2 ) } = \langle - 3 u \cdot 2 \rangle \oplus \langle u ^ { \prime } \cdot 2 \rangle \oplus U _ { 1 } \oplus V _ { 1 }\) with discriminant \(- 3 u u ^ { \prime } . 2 ^ { 6 }\) , respectively \(u u ^ { \prime } { \cdot } 2 ^ { 6 }\) . In the first case \(L = L _ { q ^ { \# } } ^ { ( 1 ) } \oplus U \oplus U\) and so we must have \(u u ^ { \prime } = 1\) , i.e, \(u = u ^ { \prime }\) . The second case is impossible.
Definition: Definition
Definition. The genus invariant of a non-degenerate integral quadratic lattice \(( L , q )\) is the triple \(\mathsf { g } ( L ) : = ( r _ { + } , r _ { - } , [ \boldsymbol { q } _ { L } ^ { \# } ] )\) , where \(( r _ { + } , r _ { - } )\) is the signature of \(L\) and \([ q _ { L } ^ { \# } ]\) is the isometry class of the discriminant quadratic form \(q _ { L } ^ { \# }\) of \(L\) .
Theorem 11.3.1
Theorem 11.3.1 (Nikulin [171, Cor. 1.9.4]). The genus of a non-degenerate quadratic lattice \(( L , q )\) is completely determined by its genus invariant \(\xi ( L )\) .
Example / comparison
The lattices \(\Gamma _ { 1 6 }\) and \(\Gamma _ { 8 } \oplus \Gamma _ { 8 } ( \simeq E _ { 8 } \oplus E _ { 8 } )\) . We calculate the Hasse invariants using Lemma 10.4.2. Being unimodular, both have \(\varepsilon _ { p } = 1\) for \(p\) odd. Since \(\mathrm { d i s c ( { \Gamma _ { 8 } } ) = \mathrm { d i s c } ( { \Gamma _ { 1 6 } } ) = 1 }\) we find \(\mu ( \Gamma _ { 8 } \oplus \Gamma _ { 8 } ) = \mu ( \Gamma _ { 1 6 } ) = - 1 6\) and so \(\varepsilon _ { 2 } ( f ) = 1\) in both cases. It follows that both lattices have the same Hasse invariants. This shows that there is an isometry between the localized lattices, but the isometry is a priori only over \(\mathbb { Q } _ { p }\) . The case \(\mathbb { Q } _ { \infty }\) is evident. The next example shows that they are isometric over \(\mathbb { Z } _ { p }\) , i.e. they belong to the same genus.
Example / comparison
Unimodular quadratic lattices. These have 0 discriminant form and hence the genus is completely determined by the signature. In particular all positive definite even unimodular lattices of the same rank are in one genus.