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Riccardo
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I'm reading Seidel's paper Graded Lagrangian submanifolds where he introduces the absolute Maslov index of a pair of graded lagrangians as follows:

Let $\mathcal{L}(V,\beta)$ be the Lagrangian Grassmannian of the symplectic vector space $(V,\beta)$ $(\dim V=2n)$. Let $\mathcal{L}^N(V,\beta)$ be the $\Bbb Z/N$ cover associated to the reduction modulo $N$ of the Maslov class $C \in H^1(\mathcal{L}(V,\beta), \Bbb Z)$. Let $\tilde{\Lambda}_0,\tilde{\Lambda}_1 \in \mathcal{L}^N(V,\beta)$ be two graded Lagrangians (more precisely $\Lambda_0,\Lambda_1$ are the two Lagrangians with chosen lifts $\tilde{\Lambda}_0,\tilde{\Lambda}_1$. Assume $\Lambda_0 \pitchfork \Lambda_1$. Choose two paths $$\tilde{\lambda}_0,\tilde{\lambda}_1 \colon [0,1]\to \mathcal{L}^N(V,\beta)$$ with the same starting point (call it $A$) and ending in $\tilde{\Lambda}_0$ and $\tilde{\Lambda}_1$ respectively.

Now the absolute Maslov index $\tilde{\mu}$ for the pair $(\tilde{\Lambda}_0,\tilde{\Lambda}_1)$ is defined by $$ \tilde{\mu}(\tilde{\Lambda}_0,\tilde{\Lambda}_1) := n/2 - \mu(\lambda_0,\lambda_1) \in \Bbb Z/N$$

where the latter is the Maslov index for a pair of paths. My understanding of the latter is that it's an half-integer, i.e it's of the form $a/2$ for $a \in \Bbb Z$.

Why is it an element of $\Bbb Z/N$ and not $\frac{1}{2}\Bbb Z/N$? in other words, why $\mu(\lambda_0,\lambda_1)$ is $\Bbb Z/N$ valued when $n$ is even and "is not integer" when $n$ is odd independently from $\lambda_0,\lambda_1$?

Since the two lagrangians are taken to be transverse, we need to prove that $ \frac{1}{2}\text{sign }\Gamma(\lambda_0,\lambda_1,0)$ (Notation took from Robbin, Salamon - The Maslov Index for pathsRobbin, Salamon - The Maslov Index for paths) is an integer when $n$ is even, half an integer otherwise. By following Robbin and Salamon's paper (page 12) we know that: \begin{align*} \Gamma(\lambda_0,\lambda_1,0) &:= \Gamma(\lambda_0,A,0)-\Gamma(\lambda_1,A,0) \end{align*}

Where $\Gamma(\lambda_0,A,0)$ is the quadratic form defined as follows: Let $W$ be a fixed Lagrangian complement for $A$ and for $v \in A$ and small $t$ define $w(t) \in W$ by $v+w(t)\in \lambda_0(t)$. Then the quadratic form $\Gamma$ is defined as $$\Gamma(v) = \left.\dfrac{d}{dt}\right\rvert_{t=0} \beta(v,w(t))$$ (The definition turns out to be independent of the choice of $W$)

The problem is that I don't know how to compute the signature for this quadratic form. Any help is appreciated.

I'm reading Seidel's paper Graded Lagrangian submanifolds where he introduces the absolute Maslov index of a pair of graded lagrangians as follows:

Let $\mathcal{L}(V,\beta)$ be the Lagrangian Grassmannian of the symplectic vector space $(V,\beta)$ $(\dim V=2n)$. Let $\mathcal{L}^N(V,\beta)$ be the $\Bbb Z/N$ cover associated to the reduction modulo $N$ of the Maslov class $C \in H^1(\mathcal{L}(V,\beta), \Bbb Z)$. Let $\tilde{\Lambda}_0,\tilde{\Lambda}_1 \in \mathcal{L}^N(V,\beta)$ be two graded Lagrangians (more precisely $\Lambda_0,\Lambda_1$ are the two Lagrangians with chosen lifts $\tilde{\Lambda}_0,\tilde{\Lambda}_1$. Assume $\Lambda_0 \pitchfork \Lambda_1$. Choose two paths $$\tilde{\lambda}_0,\tilde{\lambda}_1 \colon [0,1]\to \mathcal{L}^N(V,\beta)$$ with the same starting point (call it $A$) and ending in $\tilde{\Lambda}_0$ and $\tilde{\Lambda}_1$ respectively.

Now the absolute Maslov index $\tilde{\mu}$ for the pair $(\tilde{\Lambda}_0,\tilde{\Lambda}_1)$ is defined by $$ \tilde{\mu}(\tilde{\Lambda}_0,\tilde{\Lambda}_1) := n/2 - \mu(\lambda_0,\lambda_1) \in \Bbb Z/N$$

where the latter is the Maslov index for a pair of paths. My understanding of the latter is that it's an half-integer, i.e it's of the form $a/2$ for $a \in \Bbb Z$.

Why is it an element of $\Bbb Z/N$ and not $\frac{1}{2}\Bbb Z/N$? in other words, why $\mu(\lambda_0,\lambda_1)$ is $\Bbb Z/N$ valued when $n$ is even and "is not integer" when $n$ is odd independently from $\lambda_0,\lambda_1$?

Since the two lagrangians are taken to be transverse, we need to prove that $ \frac{1}{2}\text{sign }\Gamma(\lambda_0,\lambda_1,0)$ (Notation took from Robbin, Salamon - The Maslov Index for paths) is an integer when $n$ is even, half an integer otherwise. By following Robbin and Salamon's paper (page 12) we know that: \begin{align*} \Gamma(\lambda_0,\lambda_1,0) &:= \Gamma(\lambda_0,A,0)-\Gamma(\lambda_1,A,0) \end{align*}

Where $\Gamma(\lambda_0,A,0)$ is the quadratic form defined as follows: Let $W$ be a fixed Lagrangian complement for $A$ and for $v \in A$ and small $t$ define $w(t) \in W$ by $v+w(t)\in \lambda_0(t)$. Then the quadratic form $\Gamma$ is defined as $$\Gamma(v) = \left.\dfrac{d}{dt}\right\rvert_{t=0} \beta(v,w(t))$$ (The definition turns out to be independent of the choice of $W$)

The problem is that I don't know how to compute the signature for this quadratic form. Any help is appreciated.

I'm reading Seidel's paper Graded Lagrangian submanifolds where he introduces the absolute Maslov index of a pair of graded lagrangians as follows:

Let $\mathcal{L}(V,\beta)$ be the Lagrangian Grassmannian of the symplectic vector space $(V,\beta)$ $(\dim V=2n)$. Let $\mathcal{L}^N(V,\beta)$ be the $\Bbb Z/N$ cover associated to the reduction modulo $N$ of the Maslov class $C \in H^1(\mathcal{L}(V,\beta), \Bbb Z)$. Let $\tilde{\Lambda}_0,\tilde{\Lambda}_1 \in \mathcal{L}^N(V,\beta)$ be two graded Lagrangians (more precisely $\Lambda_0,\Lambda_1$ are the two Lagrangians with chosen lifts $\tilde{\Lambda}_0,\tilde{\Lambda}_1$. Assume $\Lambda_0 \pitchfork \Lambda_1$. Choose two paths $$\tilde{\lambda}_0,\tilde{\lambda}_1 \colon [0,1]\to \mathcal{L}^N(V,\beta)$$ with the same starting point (call it $A$) and ending in $\tilde{\Lambda}_0$ and $\tilde{\Lambda}_1$ respectively.

Now the absolute Maslov index $\tilde{\mu}$ for the pair $(\tilde{\Lambda}_0,\tilde{\Lambda}_1)$ is defined by $$ \tilde{\mu}(\tilde{\Lambda}_0,\tilde{\Lambda}_1) := n/2 - \mu(\lambda_0,\lambda_1) \in \Bbb Z/N$$

where the latter is the Maslov index for a pair of paths. My understanding of the latter is that it's an half-integer, i.e it's of the form $a/2$ for $a \in \Bbb Z$.

Why is it an element of $\Bbb Z/N$ and not $\frac{1}{2}\Bbb Z/N$? in other words, why $\mu(\lambda_0,\lambda_1)$ is $\Bbb Z/N$ valued when $n$ is even and "is not integer" when $n$ is odd independently from $\lambda_0,\lambda_1$?

Since the two lagrangians are taken to be transverse, we need to prove that $ \frac{1}{2}\text{sign }\Gamma(\lambda_0,\lambda_1,0)$ (Notation took from Robbin, Salamon - The Maslov Index for paths) is an integer when $n$ is even, half an integer otherwise. By following Robbin and Salamon's paper (page 12) we know that: \begin{align*} \Gamma(\lambda_0,\lambda_1,0) &:= \Gamma(\lambda_0,A,0)-\Gamma(\lambda_1,A,0) \end{align*}

Where $\Gamma(\lambda_0,A,0)$ is the quadratic form defined as follows: Let $W$ be a fixed Lagrangian complement for $A$ and for $v \in A$ and small $t$ define $w(t) \in W$ by $v+w(t)\in \lambda_0(t)$. Then the quadratic form $\Gamma$ is defined as $$\Gamma(v) = \left.\dfrac{d}{dt}\right\rvert_{t=0} \beta(v,w(t))$$ (The definition turns out to be independent of the choice of $W$)

The problem is that I don't know how to compute the signature for this quadratic form. Any help is appreciated.

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Riccardo
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  • 19

A clarification in the definition of Seidel's absolute Maslov index for a pair of transverse Lagrangians

I'm reading Seidel's paper Graded Lagrangian submanifolds where he introduces the absolute Maslov index of a pair of graded lagrangians as follows:

Let $\mathcal{L}(V,\beta)$ be the Lagrangian Grassmannian of the symplectic vector space $(V,\beta)$ $(\dim V=2n)$. Let $\mathcal{L}^N(V,\beta)$ be the $\Bbb Z/N$ cover associated to the reduction modulo $N$ of the Maslov class $C \in H^1(\mathcal{L}(V,\beta), \Bbb Z)$. Let $\tilde{\Lambda}_0,\tilde{\Lambda}_1 \in \mathcal{L}^N(V,\beta)$ be two graded Lagrangians (more precisely $\Lambda_0,\Lambda_1$ are the two Lagrangians with chosen lifts $\tilde{\Lambda}_0,\tilde{\Lambda}_1$. Assume $\Lambda_0 \pitchfork \Lambda_1$. Choose two paths $$\tilde{\lambda}_0,\tilde{\lambda}_1 \colon [0,1]\to \mathcal{L}^N(V,\beta)$$ with the same starting point (call it $A$) and ending in $\tilde{\Lambda}_0$ and $\tilde{\Lambda}_1$ respectively.

Now the absolute Maslov index $\tilde{\mu}$ for the pair $(\tilde{\Lambda}_0,\tilde{\Lambda}_1)$ is defined by $$ \tilde{\mu}(\tilde{\Lambda}_0,\tilde{\Lambda}_1) := n/2 - \mu(\lambda_0,\lambda_1) \in \Bbb Z/N$$

where the latter is the Maslov index for a pair of paths. My understanding of the latter is that it's an half-integer, i.e it's of the form $a/2$ for $a \in \Bbb Z$.

Why is it an element of $\Bbb Z/N$ and not $\frac{1}{2}\Bbb Z/N$? in other words, why $\mu(\lambda_0,\lambda_1)$ is $\Bbb Z/N$ valued when $n$ is even and "is not integer" when $n$ is odd independently from $\lambda_0,\lambda_1$?

Since the two lagrangians are taken to be transverse, we need to prove that $ \frac{1}{2}\text{sign }\Gamma(\lambda_0,\lambda_1,0)$ (Notation took from Robbin, Salamon - The Maslov Index for paths) is an integer when $n$ is even, half an integer otherwise. By following Robbin and Salamon's paper (page 12) we know that: \begin{align*} \Gamma(\lambda_0,\lambda_1,0) &:= \Gamma(\lambda_0,A,0)-\Gamma(\lambda_1,A,0) \end{align*}

Where $\Gamma(\lambda_0,A,0)$ is the quadratic form defined as follows: Let $W$ be a fixed Lagrangian complement for $A$ and for $v \in A$ and small $t$ define $w(t) \in W$ by $v+w(t)\in \lambda_0(t)$. Then the quadratic form $\Gamma$ is defined as $$\Gamma(v) = \left.\dfrac{d}{dt}\right\rvert_{t=0} \beta(v,w(t))$$ (The definition turns out to be independent of the choice of $W$)

The problem is that I don't know how to compute the signature for this quadratic form. Any help is appreciated.