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If $M_{g}$ is the moduli space of Riemann surfacessurface of genus $g$, and $M^1_{g}$ is the moduli space of Riemann surfaceasurface of genus $g$ with one boundary, how can we show that the natural map:

$M^1_{g} \rightarrow M_{g}$

is a fibration (I think it is a fiber bundle), and the fiber over a point $S \in M_{g}$ is the sphere bundle associated to the tangent bundle of the Riemann surface $S$?

I don't have appropriate coordinates;coordinates that is probably why I can't show it.

I am asking this question because I want to show that if $M^{m,b}_{g,n}$ is the moduli space of Riemann surfacessurfacw of genus $g$ and $b$ boundary components, and with $n$ interior marked points, $m=(m_1,m_2,\ldots,m_{b})$ marked points on the boundary, $m_i$ marked points on the $i$th boundary component, then the homology dimension

$H_i(M^{m,b}_{g,n},Q)=0,$ for $i\ge 6g-7+2n+3b+m$ possibly excludes some bad low dimensional cases.

I learned one way is totp assoicate it to the mapping class group (especially we havegot a virtual cohomological dimension of the mapping class group by Harer, "OnHarer"on virtual cohomological dimension of mapping class group of oriented surface") and the homology of mapping class group is the same as the homology of the above moduli space, but I still don't understand how it works. Thank you for help.

If $M_{g}$ is the moduli space of Riemann surfaces of genus $g$, and $M^1_{g}$ is the moduli space of Riemann surfacea of genus $g$ with one boundary, how can we show that the natural map:

$M^1_{g} \rightarrow M_{g}$

is a fibration (I think it is a fiber bundle), and the fiber over a point $S \in M_{g}$ is the sphere bundle associated to the tangent bundle of the Riemann surface $S$?

I don't have appropriate coordinates; that is probably why I can't show it.

I am asking this question because I want to show that if $M^{m,b}_{g,n}$ is the moduli space of Riemann surfaces of genus $g$ and $b$ boundary components, and with $n$ interior marked points, $m=(m_1,m_2,\ldots,m_{b})$ marked points on the boundary, $m_i$ marked points on the $i$th boundary component, then the homology dimension

$H_i(M^{m,b}_{g,n},Q)=0,$ for $i\ge 6g-7+2n+3b+m$ possibly excludes some bad low dimensional cases.

I learned one way is to assoicate it to the mapping class group (especially we have a virtual cohomological dimension of the mapping class group by Harer, "On virtual cohomological dimension of mapping class group of oriented surface") and the homology of mapping class group is the same as the homology of the above moduli space, but I still don't understand how it works. Thank you for help.

If $M_{g}$ is the moduli space of Riemann surface of genus $g$, $M^1_{g}$ is the moduli space of Riemann surface of genus $g$ with one boundary, how can we show that the natural map:

$M^1_{g} \rightarrow M_{g}$

the fiber over a point $S \in M_{g}$ is the sphere bundle associated to the tangent bundle of Riemann surface $S$?

I don't have appropriate coordinates that is probably why I can't show it.

I am asking this question because I want to show that if $M^{m,b}_{g,n}$ is the moduli space of Riemann surfacw of genus $g$ and $b$ boundary components, and with $n$ interior marked points, $m=(m_1,m_2,\ldots,m_{b})$ marked points on the boundary, $m_i$ marked points on the $i$th boundary component, then the homology dimension

$H_i(M^{m,b}_{g,n},Q)=0,$ for $i\ge 6g-7+2n+3b+m$ possibly excludes some bad low dimensional cases.

I learned one way is tp assoicate it to mapping class group (especially we got a virtual cohomological dimension of mapping class group by Harer"on virtual cohomological dimension of mapping class group of oriented surface") and homology of mapping class group is the same as homology of the above moduli space, but I still don't understand how it works. Thank you for help.

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Charles Matthews
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a question Question related to modulthe moduli space of Riemann surfacesurfaces and a fibration

If $M_{g}$ is the moduli space of Riemann surfacesurfaces of genus $g$, and $M^1_{g}$ is the moduli space of Riemann surfacesurfacea of genus $g$ with one boundary, how can we show that the natural map:

$M^1_{g} \rightarrow M_{g}$

is a fibration (I think it is a fiber bundle), and the fiber over a point $S \in M_{g}$ is the sphere bundle associated to the tangent bundle of the Riemann surface $S$?

I don't have appropriate coordinatescoordinates; that is probably why I can't show it.

I am asking this question because I want to show that if $M^{m,b}_{g,n}$ is the moduli space of Riemann surfacwsurfaces of genus $g$ and $b$ boundary components, and with $n$ interior marked points, $m=(m_1,m_2,\ldots,m_{b})$ marked points on the boundary, $m_i$ marked points on the $i$th boundary component, then the homology dimension

$H_i(M^{m,b}_{g,n},Q)=0,$ for $i\ge 6g-7+2n+3b+m$ possibly excludes some bad low dimensional cases.

I learned one way is tpto assoicate it to the mapping class group (especially we gothave a virtual cohomological dimension of the mapping class group by Harer"onHarer, "On virtual cohomological dimension of mapping class group of oriented surface") and the homology of mapping class group is the same as the homology of the above moduli space, but I still don't understand how it works. Thank you for help.

a question related to modul space of Riemann surface and fibration

If $M_{g}$ is the moduli space of Riemann surface of genus $g$, $M^1_{g}$ is the moduli space of Riemann surface of genus $g$ with one boundary, how can we show that the natural map:

$M^1_{g} \rightarrow M_{g}$

is a fibration (I think it is a fiber bundle), and the fiber over a point $S \in M_{g}$ is the sphere bundle associated to the tangent bundle of Riemann surface $S$?

I don't have appropriate coordinates that is probably why I can't show it.

I am asking this question because I want to show that if $M^{m,b}_{g,n}$ is the moduli space of Riemann surfacw of genus $g$ and $b$ boundary components, and with $n$ interior marked points, $m=(m_1,m_2,\ldots,m_{b})$ marked points on the boundary, $m_i$ marked points on the $i$th boundary component, then the homology dimension

$H_i(M^{m,b}_{g,n},Q)=0,$ for $i\ge 6g-7+2n+3b+m$ possibly excludes some bad low dimensional cases.

I learned one way is tp assoicate it to mapping class group (especially we got a virtual cohomological dimension of mapping class group by Harer"on virtual cohomological dimension of mapping class group of oriented surface") and homology of mapping class group is the same as homology of the above moduli space, but I still don't understand how it works. Thank you for help.

Question related to the moduli space of Riemann surfaces and a fibration

If $M_{g}$ is the moduli space of Riemann surfaces of genus $g$, and $M^1_{g}$ is the moduli space of Riemann surfacea of genus $g$ with one boundary, how can we show that the natural map:

$M^1_{g} \rightarrow M_{g}$

is a fibration (I think it is a fiber bundle), and the fiber over a point $S \in M_{g}$ is the sphere bundle associated to the tangent bundle of the Riemann surface $S$?

I don't have appropriate coordinates; that is probably why I can't show it.

I am asking this question because I want to show that if $M^{m,b}_{g,n}$ is the moduli space of Riemann surfaces of genus $g$ and $b$ boundary components, and with $n$ interior marked points, $m=(m_1,m_2,\ldots,m_{b})$ marked points on the boundary, $m_i$ marked points on the $i$th boundary component, then the homology dimension

$H_i(M^{m,b}_{g,n},Q)=0,$ for $i\ge 6g-7+2n+3b+m$ possibly excludes some bad low dimensional cases.

I learned one way is to assoicate it to the mapping class group (especially we have a virtual cohomological dimension of the mapping class group by Harer, "On virtual cohomological dimension of mapping class group of oriented surface") and the homology of mapping class group is the same as the homology of the above moduli space, but I still don't understand how it works. Thank you for help.

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HYYY
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If $M_{g}$ is the moduli space of Riemann surface of genus $g$, $M^1_{g}$ is the moduli space of Riemann surface of genus $g$ with one boundary, how can we show that the natural map:

$M^1_{g} \rightarrow M_{g}$

is a fibration (I think it is a fiber bundle), and the fiber over a point $S \in M_{g}$ is the sphere bundle associated to the tangent bundle of Riemann surface $S$?

I don't have appropriate coordinates that is probably why I can't show it.

I am asking this question because I want to show that if $M^{m,b}_{g,n}$ is the moduli space of Riemann surfacw of genus $g$ and $b$ boundary components, and with $n$ interior marked points, $m=(m_1,m_2,\ldots,m_{b})$ marked points on the boundary, $m_i$ marked points on the $i$th boundary component, then the homology dimension

$H_i(M^{m,b}_{g,n},Q)=0,$ for $i\ge 6g-7+2n+3b+m$ possibly excludes some bad low dimensional cases.

I learned one way is tp assoicate it to mapping class group (especially we got a virtual cohomological dimension of mapping class group by Harer"on virtual cohomological dimension of mapping class group of oriented surface") and homology of mapping class group is the same as homology of the above moduli space, but I still don't understand how it works. Thank you for help.

If $M_{g}$ is the moduli space of Riemann surface of genus $g$, $M^1_{g}$ is the moduli space of Riemann surface of genus $g$ with one boundary, how can we show that the natural map:

$M^1_{g} \rightarrow M_{g}$

is a fibration (I think it is a fiber bundle), and the fiber over a point $S \in M_{g}$ is the sphere bundle associated to the tangent bundle of Riemann surface $S$?

because I want to show that if $M^{m,b}_{g,n}$ is the moduli space of Riemann surfacw of genus $g$ and $b$ boundary components, and with $n$ interior marked points, $m=(m_1,m_2,\ldots,m_{b})$ marked points on the boundary, $m_i$ marked points on the $i$th boundary component, then the homology dimension

$H_i(M^{m,b}_{g,n},Q)=0,$ for $i\ge 6g-7+2n+3b+m$ possibly excludes some bad low dimensional cases.

I learned one way is tp assoicate it to mapping class group (especially we got a virtual cohomological dimension of mapping class group by Harer"on virtual cohomological dimension of mapping class group of oriented surface") and homology of mapping class group is the same as homology of the above moduli space, but I still don't understand how it works. Thank you for help.

If $M_{g}$ is the moduli space of Riemann surface of genus $g$, $M^1_{g}$ is the moduli space of Riemann surface of genus $g$ with one boundary, how can we show that the natural map:

$M^1_{g} \rightarrow M_{g}$

is a fibration (I think it is a fiber bundle), and the fiber over a point $S \in M_{g}$ is the sphere bundle associated to the tangent bundle of Riemann surface $S$?

I don't have appropriate coordinates that is probably why I can't show it.

I am asking this question because I want to show that if $M^{m,b}_{g,n}$ is the moduli space of Riemann surfacw of genus $g$ and $b$ boundary components, and with $n$ interior marked points, $m=(m_1,m_2,\ldots,m_{b})$ marked points on the boundary, $m_i$ marked points on the $i$th boundary component, then the homology dimension

$H_i(M^{m,b}_{g,n},Q)=0,$ for $i\ge 6g-7+2n+3b+m$ possibly excludes some bad low dimensional cases.

I learned one way is tp assoicate it to mapping class group (especially we got a virtual cohomological dimension of mapping class group by Harer"on virtual cohomological dimension of mapping class group of oriented surface") and homology of mapping class group is the same as homology of the above moduli space, but I still don't understand how it works. Thank you for help.

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HYYY
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