Explicit Riemann Hilbert correspondence
$begingroup$
For simplicity, we assume that $X=mathbb P_{mathbb C}^1-{s_1, s_2, dots, s_k}$ and $infty in X$.
Consider the trivial bundle $E=mathcal O_X^r$ with the connection $nabla$ induced by a Fushcian system on $X$, i.e.
$$nabla:= d+sum_{i=1}^k frac{A_i}{z-s_i},$$
where $A_i$'s are $rtimes r$ constant matrices over $mathbb C$ such that $sum_{i=1}^kA_i=0$.
Then we have a flat bundle with connection (or a logarithmic connection), note it only has regular singularities.
My question is that what is the corresponding monodromy representation to $(E,nabla)$ via the Riemann Hilbert correspondence?
The known part is that locally at each $z=s_i$, the monodromy $T_i$ can be obtained by the residue map: $T_i=exp(2pi i A_i)$, but this seems not to admit a unique representation up to isomorphism.
Is there any more properties/facts around the question?
connections d-modules monodromy local-systems
asked Feb 24 at 10:36
LongmaLongma
384
$endgroup$
add a comment |
$begingroup$
For simplicity, we assume that $X=mathbb P_{mathbb C}^1-{s_1, s_2, dots, s_k}$ and $infty in X$.
Consider the trivial bundle $E=mathcal O_X^r$ with the connection $nabla$ induced by a Fushcian system on $X$, i.e.
$$nabla:= d+sum_{i=1}^k frac{A_i}{z-s_i},$$
where $A_i$'s are $rtimes r$ constant matrices over $mathbb C$ such that $sum_{i=1}^kA_i=0$.
Then we have a flat bundle with connection (or a logarithmic connection), note it only has regular singularities.
My question is that what is the corresponding monodromy representation to $(E,nabla)$ via the Riemann Hilbert correspondence?
The known part is that locally at each $z=s_i$, the monodromy $T_i$ can be obtained by the residue map: $T_i=exp(2pi i A_i)$, but this seems not to admit a unique representation up to isomorphism.
Is there any more properties/facts around the question?
connections d-modules monodromy local-systems
asked Feb 24 at 10:36
LongmaLongma
384
$endgroup$
1
$begingroup$
You cannot solve ode "explicitly" - so cannot find explicit monodromies. Rare exceptions like Knizhnik-Zamolodchikov equation are due some hidden Lie group symmetry inside.
$endgroup$
– Alexander Chervov
Feb 24 at 12:20
$begingroup$
By Riemann Hilbert correspondence, if the sum of A_i is zero, and fix a map $mathbb C/mathbb Zto mathbb C$, then there exists a unique local system, namely a unique representation.
$endgroup$
– Longma
Feb 25 at 9:06
add a comment |
$begingroup$
For simplicity, we assume that $X=mathbb P_{mathbb C}^1-{s_1, s_2, dots, s_k}$ and $infty in X$.
Consider the trivial bundle $E=mathcal O_X^r$ with the connection $nabla$ induced by a Fushcian system on $X$, i.e.
$$nabla:= d+sum_{i=1}^k frac{A_i}{z-s_i},$$
where $A_i$'s are $rtimes r$ constant matrices over $mathbb C$ such that $sum_{i=1}^kA_i=0$.
Then we have a flat bundle with connection (or a logarithmic connection), note it only has regular singularities.
My question is that what is the corresponding monodromy representation to $(E,nabla)$ via the Riemann Hilbert correspondence?
The known part is that locally at each $z=s_i$, the monodromy $T_i$ can be obtained by the residue map: $T_i=exp(2pi i A_i)$, but this seems not to admit a unique representation up to isomorphism.
Is there any more properties/facts around the question?
connections d-modules monodromy local-systems
asked Feb 24 at 10:36
LongmaLongma
384
$endgroup$
For simplicity, we assume that $X=mathbb P_{mathbb C}^1-{s_1, s_2, dots, s_k}$ and $infty in X$.
Consider the trivial bundle $E=mathcal O_X^r$ with the connection $nabla$ induced by a Fushcian system on $X$, i.e.
$$nabla:= d+sum_{i=1}^k frac{A_i}{z-s_i},$$
where $A_i$'s are $rtimes r$ constant matrices over $mathbb C$ such that $sum_{i=1}^kA_i=0$.
Then we have a flat bundle with connection (or a logarithmic connection), note it only has regular singularities.
My question is that what is the corresponding monodromy representation to $(E,nabla)$ via the Riemann Hilbert correspondence?
The known part is that locally at each $z=s_i$, the monodromy $T_i$ can be obtained by the residue map: $T_i=exp(2pi i A_i)$, but this seems not to admit a unique representation up to isomorphism.
Is there any more properties/facts around the question?
connections d-modules monodromy local-systems
connections d-modules monodromy local-systems
asked Feb 24 at 10:36
LongmaLongma
384
asked Feb 24 at 10:36
LongmaLongma
384
edited Feb 25 at 9:07
Longma
asked Feb 24 at 10:36
LongmaLongma
384
asked Feb 24 at 10:36
LongmaLongma
384
asked Feb 24 at 10:36
LongmaLongma
384
384
1
$begingroup$
You cannot solve ode "explicitly" - so cannot find explicit monodromies. Rare exceptions like Knizhnik-Zamolodchikov equation are due some hidden Lie group symmetry inside.
$endgroup$
– Alexander Chervov
Feb 24 at 12:20
$begingroup$
By Riemann Hilbert correspondence, if the sum of A_i is zero, and fix a map $mathbb C/mathbb Zto mathbb C$, then there exists a unique local system, namely a unique representation.
$endgroup$
– Longma
Feb 25 at 9:06
add a comment |
1
$begingroup$
You cannot solve ode "explicitly" - so cannot find explicit monodromies. Rare exceptions like Knizhnik-Zamolodchikov equation are due some hidden Lie group symmetry inside.
$endgroup$
– Alexander Chervov
Feb 24 at 12:20
$begingroup$
By Riemann Hilbert correspondence, if the sum of A_i is zero, and fix a map $mathbb C/mathbb Zto mathbb C$, then there exists a unique local system, namely a unique representation.
$endgroup$
– Longma
Feb 25 at 9:06
1
1
$begingroup$
You cannot solve ode "explicitly" - so cannot find explicit monodromies. Rare exceptions like Knizhnik-Zamolodchikov equation are due some hidden Lie group symmetry inside.
$endgroup$
– Alexander Chervov
Feb 24 at 12:20
$begingroup$
You cannot solve ode "explicitly" - so cannot find explicit monodromies. Rare exceptions like Knizhnik-Zamolodchikov equation are due some hidden Lie group symmetry inside.
$endgroup$
– Alexander Chervov
Feb 24 at 12:20
$begingroup$
By Riemann Hilbert correspondence, if the sum of A_i is zero, and fix a map $mathbb C/mathbb Zto mathbb C$, then there exists a unique local system, namely a unique representation.
$endgroup$
– Longma
Feb 25 at 9:06
$begingroup$
By Riemann Hilbert correspondence, if the sum of A_i is zero, and fix a map $mathbb C/mathbb Zto mathbb C$, then there exists a unique local system, namely a unique representation.
$endgroup$
– Longma
Feb 25 at 9:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It depends on what is called "explicit". If $k>2$, monodromy representation is a transcendental function of the $A_j$ and $s_j$. When $d=0$, it was expressed as an everywhere convergent power series in
the $A_j$ whose coefficients are explicit (rational) functions in $s_j$ by Lappo-Danilevski:
Lappo-Danilevsky, J. A. Mémoires sur la théorie des systèmes des équations différentielles linéaires. (French) Chelsea Publishing Co., New York, N. Y., 1953.
If $dneq 0$, the system is not Fuchsian: singularity at $infty$ is irregular. If $k=2, dneq 0$ your system already contains the
"prolate/oblate spheroid equations", which were studied much and no reasonable explicit formula for the monodromy
is known. There are asymptotics, of course. A recent paper about this special case, with a good reference list is
Richard-Jung, F.; Ramis, J.-P.; Thomann, J.; Fauvet, F. New characterizations for the eigenvalues of the prolate spheroidal wave equation. Stud. Appl. Math. 138 (2017), no. 1, 3–42.
The case $d=0$, $k=3$ was also much studied. The papers usually refer to "Heun's equation".
answered Feb 24 at 14:28
Alexandre EremenkoAlexandre Eremenko
50.5k6140257
$endgroup$
1
$begingroup$
In physics they call it 'time ordered exponent' en.m.wikipedia.org/wiki/Ordered_exponential
$endgroup$
– Alexander Chervov
Feb 24 at 14:35
$begingroup$
@Alexander Chervov: good remark! This gives many references which are more modern than the original papers of Lappo-Danilevsky.
$endgroup$
– Alexandre Eremenko
Feb 24 at 14:38
$begingroup$
@Alexander Chervov: what is this m.wikipedia? I could not find any general description.
$endgroup$
– Alexandre Eremenko
Feb 24 at 23:33
$begingroup$
M stands for mobile it is the same wiki, but it looks like that from mobile devices
$endgroup$
– Alexander Chervov
Feb 25 at 4:28
$begingroup$
Thanks a lot for your help. Maybe I should assume that the sum of A_i is 0 and d is not 0 so that the system is Fuchsian, do you have any idea in this case?
$endgroup$
– Longma
Feb 25 at 9:03
|
show 1 more comment
$begingroup$
You are correct that the monodromy representation is given by the $T_i$. To address your concerns about this being unique up to isomorphism, notice that a change of basis of $mathcal{O}_X^r$ induces a corresponding (compatible) change to the $T_i$.
You might be interested in the following references: Sections 5.1.1 and 5.1.2 of Hotta, Tanasaki, and Takeuchi's "D-Modules, Perverse Sheaves, and Representation Theory"; and Chapter III of "Algebraic D-Modules" by Borel, et al.
answered Feb 24 at 17:41
Avi SteinerAvi Steiner
1,52211229
$endgroup$
$begingroup$
Thanks a lot. Actually the reference you gave is exactly the book I am reading. In this book, they gave a way to obtain the monodromy representation, however, their method requires that all monodromy are commuting invertible matrices, which I do not get the point, do you have any idea?
$endgroup$
– Longma
Feb 25 at 9:03
$begingroup$
@Longma if you don’t have commuting matrices, you don’t get an integrable connection, and in particular can’t use the Riemann-Hilbert Correspondence. In fact, you won’t even get a D-module
$endgroup$
– Avi Steiner
Feb 25 at 16:03
$begingroup$
In higher dimensional cases, I totally agree with you. However, in my example, we are working on the one dimensional case, namely, every two components of the NCD divisor has no intersection, so we don't need to require that the monodromy commutates. (Since the commutativity condition is due to monodromy agrees on the intersection of each "singularity". Does it make any sense for you?
$endgroup$
– Longma
Feb 26 at 11:04
$begingroup$
@Longma Oh! Shoot! You’re right! Yeah, there’s no particular reason you need commuting matrices. My answer still holds , though.
$endgroup$
– Avi Steiner
Feb 26 at 14:21
$begingroup$
Ok. I need more information to determine the induced map between isomorphic classes. Thanks anyway.
$endgroup$
– Longma
Feb 26 at 19:42
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It depends on what is called "explicit". If $k>2$, monodromy representation is a transcendental function of the $A_j$ and $s_j$. When $d=0$, it was expressed as an everywhere convergent power series in
the $A_j$ whose coefficients are explicit (rational) functions in $s_j$ by Lappo-Danilevski:
Lappo-Danilevsky, J. A. Mémoires sur la théorie des systèmes des équations différentielles linéaires. (French) Chelsea Publishing Co., New York, N. Y., 1953.
If $dneq 0$, the system is not Fuchsian: singularity at $infty$ is irregular. If $k=2, dneq 0$ your system already contains the
"prolate/oblate spheroid equations", which were studied much and no reasonable explicit formula for the monodromy
is known. There are asymptotics, of course. A recent paper about this special case, with a good reference list is
Richard-Jung, F.; Ramis, J.-P.; Thomann, J.; Fauvet, F. New characterizations for the eigenvalues of the prolate spheroidal wave equation. Stud. Appl. Math. 138 (2017), no. 1, 3–42.
The case $d=0$, $k=3$ was also much studied. The papers usually refer to "Heun's equation".
answered Feb 24 at 14:28
Alexandre EremenkoAlexandre Eremenko
50.5k6140257
$endgroup$
1
$begingroup$
In physics they call it 'time ordered exponent' en.m.wikipedia.org/wiki/Ordered_exponential
$endgroup$
– Alexander Chervov
Feb 24 at 14:35
$begingroup$
@Alexander Chervov: good remark! This gives many references which are more modern than the original papers of Lappo-Danilevsky.
$endgroup$
– Alexandre Eremenko
Feb 24 at 14:38
$begingroup$
@Alexander Chervov: what is this m.wikipedia? I could not find any general description.
$endgroup$
– Alexandre Eremenko
Feb 24 at 23:33
$begingroup$
M stands for mobile it is the same wiki, but it looks like that from mobile devices
$endgroup$
– Alexander Chervov
Feb 25 at 4:28
$begingroup$
Thanks a lot for your help. Maybe I should assume that the sum of A_i is 0 and d is not 0 so that the system is Fuchsian, do you have any idea in this case?
$endgroup$
– Longma
Feb 25 at 9:03
|
show 1 more comment
$begingroup$
It depends on what is called "explicit". If $k>2$, monodromy representation is a transcendental function of the $A_j$ and $s_j$. When $d=0$, it was expressed as an everywhere convergent power series in
the $A_j$ whose coefficients are explicit (rational) functions in $s_j$ by Lappo-Danilevski:
Lappo-Danilevsky, J. A. Mémoires sur la théorie des systèmes des équations différentielles linéaires. (French) Chelsea Publishing Co., New York, N. Y., 1953.
If $dneq 0$, the system is not Fuchsian: singularity at $infty$ is irregular. If $k=2, dneq 0$ your system already contains the
"prolate/oblate spheroid equations", which were studied much and no reasonable explicit formula for the monodromy
is known. There are asymptotics, of course. A recent paper about this special case, with a good reference list is
Richard-Jung, F.; Ramis, J.-P.; Thomann, J.; Fauvet, F. New characterizations for the eigenvalues of the prolate spheroidal wave equation. Stud. Appl. Math. 138 (2017), no. 1, 3–42.
The case $d=0$, $k=3$ was also much studied. The papers usually refer to "Heun's equation".
answered Feb 24 at 14:28
Alexandre EremenkoAlexandre Eremenko
50.5k6140257
$endgroup$
1
$begingroup$
In physics they call it 'time ordered exponent' en.m.wikipedia.org/wiki/Ordered_exponential
$endgroup$
– Alexander Chervov
Feb 24 at 14:35
$begingroup$
@Alexander Chervov: good remark! This gives many references which are more modern than the original papers of Lappo-Danilevsky.
$endgroup$
– Alexandre Eremenko
Feb 24 at 14:38
$begingroup$
@Alexander Chervov: what is this m.wikipedia? I could not find any general description.
$endgroup$
– Alexandre Eremenko
Feb 24 at 23:33
$begingroup$
M stands for mobile it is the same wiki, but it looks like that from mobile devices
$endgroup$
– Alexander Chervov
Feb 25 at 4:28
$begingroup$
Thanks a lot for your help. Maybe I should assume that the sum of A_i is 0 and d is not 0 so that the system is Fuchsian, do you have any idea in this case?
$endgroup$
– Longma
Feb 25 at 9:03
|
show 1 more comment
$begingroup$
It depends on what is called "explicit". If $k>2$, monodromy representation is a transcendental function of the $A_j$ and $s_j$. When $d=0$, it was expressed as an everywhere convergent power series in
the $A_j$ whose coefficients are explicit (rational) functions in $s_j$ by Lappo-Danilevski:
Lappo-Danilevsky, J. A. Mémoires sur la théorie des systèmes des équations différentielles linéaires. (French) Chelsea Publishing Co., New York, N. Y., 1953.
If $dneq 0$, the system is not Fuchsian: singularity at $infty$ is irregular. If $k=2, dneq 0$ your system already contains the
"prolate/oblate spheroid equations", which were studied much and no reasonable explicit formula for the monodromy
is known. There are asymptotics, of course. A recent paper about this special case, with a good reference list is
Richard-Jung, F.; Ramis, J.-P.; Thomann, J.; Fauvet, F. New characterizations for the eigenvalues of the prolate spheroidal wave equation. Stud. Appl. Math. 138 (2017), no. 1, 3–42.
The case $d=0$, $k=3$ was also much studied. The papers usually refer to "Heun's equation".
answered Feb 24 at 14:28
Alexandre EremenkoAlexandre Eremenko
50.5k6140257
$endgroup$
It depends on what is called "explicit". If $k>2$, monodromy representation is a transcendental function of the $A_j$ and $s_j$. When $d=0$, it was expressed as an everywhere convergent power series in
the $A_j$ whose coefficients are explicit (rational) functions in $s_j$ by Lappo-Danilevski:
Lappo-Danilevsky, J. A. Mémoires sur la théorie des systèmes des équations différentielles linéaires. (French) Chelsea Publishing Co., New York, N. Y., 1953.
If $dneq 0$, the system is not Fuchsian: singularity at $infty$ is irregular. If $k=2, dneq 0$ your system already contains the
"prolate/oblate spheroid equations", which were studied much and no reasonable explicit formula for the monodromy
is known. There are asymptotics, of course. A recent paper about this special case, with a good reference list is
Richard-Jung, F.; Ramis, J.-P.; Thomann, J.; Fauvet, F. New characterizations for the eigenvalues of the prolate spheroidal wave equation. Stud. Appl. Math. 138 (2017), no. 1, 3–42.
The case $d=0$, $k=3$ was also much studied. The papers usually refer to "Heun's equation".
answered Feb 24 at 14:28
Alexandre EremenkoAlexandre Eremenko
50.5k6140257
edited Feb 24 at 14:49
answered Feb 24 at 14:28
Alexandre EremenkoAlexandre Eremenko
50.5k6140257
answered Feb 24 at 14:28
Alexandre EremenkoAlexandre Eremenko
50.5k6140257
answered Feb 24 at 14:28
Alexandre EremenkoAlexandre Eremenko
50.5k6140257
50.5k6140257
1
$begingroup$
In physics they call it 'time ordered exponent' en.m.wikipedia.org/wiki/Ordered_exponential
$endgroup$
– Alexander Chervov
Feb 24 at 14:35
$begingroup$
@Alexander Chervov: good remark! This gives many references which are more modern than the original papers of Lappo-Danilevsky.
$endgroup$
– Alexandre Eremenko
Feb 24 at 14:38
$begingroup$
@Alexander Chervov: what is this m.wikipedia? I could not find any general description.
$endgroup$
– Alexandre Eremenko
Feb 24 at 23:33
$begingroup$
M stands for mobile it is the same wiki, but it looks like that from mobile devices
$endgroup$
– Alexander Chervov
Feb 25 at 4:28
$begingroup$
Thanks a lot for your help. Maybe I should assume that the sum of A_i is 0 and d is not 0 so that the system is Fuchsian, do you have any idea in this case?
$endgroup$
– Longma
Feb 25 at 9:03
|
show 1 more comment
1
$begingroup$
In physics they call it 'time ordered exponent' en.m.wikipedia.org/wiki/Ordered_exponential
$endgroup$
– Alexander Chervov
Feb 24 at 14:35
$begingroup$
@Alexander Chervov: good remark! This gives many references which are more modern than the original papers of Lappo-Danilevsky.
$endgroup$
– Alexandre Eremenko
Feb 24 at 14:38
$begingroup$
@Alexander Chervov: what is this m.wikipedia? I could not find any general description.
$endgroup$
– Alexandre Eremenko
Feb 24 at 23:33
$begingroup$
M stands for mobile it is the same wiki, but it looks like that from mobile devices
$endgroup$
– Alexander Chervov
Feb 25 at 4:28
$begingroup$
Thanks a lot for your help. Maybe I should assume that the sum of A_i is 0 and d is not 0 so that the system is Fuchsian, do you have any idea in this case?
$endgroup$
– Longma
Feb 25 at 9:03
1
1
$begingroup$
In physics they call it 'time ordered exponent' en.m.wikipedia.org/wiki/Ordered_exponential
$endgroup$
– Alexander Chervov
Feb 24 at 14:35
$begingroup$
In physics they call it 'time ordered exponent' en.m.wikipedia.org/wiki/Ordered_exponential
$endgroup$
– Alexander Chervov
Feb 24 at 14:35
$begingroup$
@Alexander Chervov: good remark! This gives many references which are more modern than the original papers of Lappo-Danilevsky.
$endgroup$
– Alexandre Eremenko
Feb 24 at 14:38
$begingroup$
@Alexander Chervov: good remark! This gives many references which are more modern than the original papers of Lappo-Danilevsky.
$endgroup$
– Alexandre Eremenko
Feb 24 at 14:38
$begingroup$
@Alexander Chervov: what is this m.wikipedia? I could not find any general description.
$endgroup$
– Alexandre Eremenko
Feb 24 at 23:33
$begingroup$
@Alexander Chervov: what is this m.wikipedia? I could not find any general description.
$endgroup$
– Alexandre Eremenko
Feb 24 at 23:33
$begingroup$
M stands for mobile it is the same wiki, but it looks like that from mobile devices
$endgroup$
– Alexander Chervov
Feb 25 at 4:28
$begingroup$
M stands for mobile it is the same wiki, but it looks like that from mobile devices
$endgroup$
– Alexander Chervov
Feb 25 at 4:28
$begingroup$
Thanks a lot for your help. Maybe I should assume that the sum of A_i is 0 and d is not 0 so that the system is Fuchsian, do you have any idea in this case?
$endgroup$
– Longma
Feb 25 at 9:03
$begingroup$
Thanks a lot for your help. Maybe I should assume that the sum of A_i is 0 and d is not 0 so that the system is Fuchsian, do you have any idea in this case?
$endgroup$
– Longma
Feb 25 at 9:03
|
show 1 more comment
$begingroup$
You are correct that the monodromy representation is given by the $T_i$. To address your concerns about this being unique up to isomorphism, notice that a change of basis of $mathcal{O}_X^r$ induces a corresponding (compatible) change to the $T_i$.
You might be interested in the following references: Sections 5.1.1 and 5.1.2 of Hotta, Tanasaki, and Takeuchi's "D-Modules, Perverse Sheaves, and Representation Theory"; and Chapter III of "Algebraic D-Modules" by Borel, et al.
answered Feb 24 at 17:41
Avi SteinerAvi Steiner
1,52211229
$endgroup$
$begingroup$
Thanks a lot. Actually the reference you gave is exactly the book I am reading. In this book, they gave a way to obtain the monodromy representation, however, their method requires that all monodromy are commuting invertible matrices, which I do not get the point, do you have any idea?
$endgroup$
– Longma
Feb 25 at 9:03
$begingroup$
@Longma if you don’t have commuting matrices, you don’t get an integrable connection, and in particular can’t use the Riemann-Hilbert Correspondence. In fact, you won’t even get a D-module
$endgroup$
– Avi Steiner
Feb 25 at 16:03
$begingroup$
In higher dimensional cases, I totally agree with you. However, in my example, we are working on the one dimensional case, namely, every two components of the NCD divisor has no intersection, so we don't need to require that the monodromy commutates. (Since the commutativity condition is due to monodromy agrees on the intersection of each "singularity". Does it make any sense for you?
$endgroup$
– Longma
Feb 26 at 11:04
$begingroup$
@Longma Oh! Shoot! You’re right! Yeah, there’s no particular reason you need commuting matrices. My answer still holds , though.
$endgroup$
– Avi Steiner
Feb 26 at 14:21
$begingroup$
Ok. I need more information to determine the induced map between isomorphic classes. Thanks anyway.
$endgroup$
– Longma
Feb 26 at 19:42
add a comment |
$begingroup$
You are correct that the monodromy representation is given by the $T_i$. To address your concerns about this being unique up to isomorphism, notice that a change of basis of $mathcal{O}_X^r$ induces a corresponding (compatible) change to the $T_i$.
You might be interested in the following references: Sections 5.1.1 and 5.1.2 of Hotta, Tanasaki, and Takeuchi's "D-Modules, Perverse Sheaves, and Representation Theory"; and Chapter III of "Algebraic D-Modules" by Borel, et al.
answered Feb 24 at 17:41
Avi SteinerAvi Steiner
1,52211229
$endgroup$
$begingroup$
Thanks a lot. Actually the reference you gave is exactly the book I am reading. In this book, they gave a way to obtain the monodromy representation, however, their method requires that all monodromy are commuting invertible matrices, which I do not get the point, do you have any idea?
$endgroup$
– Longma
Feb 25 at 9:03
$begingroup$
@Longma if you don’t have commuting matrices, you don’t get an integrable connection, and in particular can’t use the Riemann-Hilbert Correspondence. In fact, you won’t even get a D-module
$endgroup$
– Avi Steiner
Feb 25 at 16:03
$begingroup$
In higher dimensional cases, I totally agree with you. However, in my example, we are working on the one dimensional case, namely, every two components of the NCD divisor has no intersection, so we don't need to require that the monodromy commutates. (Since the commutativity condition is due to monodromy agrees on the intersection of each "singularity". Does it make any sense for you?
$endgroup$
– Longma
Feb 26 at 11:04
$begingroup$
@Longma Oh! Shoot! You’re right! Yeah, there’s no particular reason you need commuting matrices. My answer still holds , though.
$endgroup$
– Avi Steiner
Feb 26 at 14:21
$begingroup$
Ok. I need more information to determine the induced map between isomorphic classes. Thanks anyway.
$endgroup$
– Longma
Feb 26 at 19:42
add a comment |
$begingroup$
You are correct that the monodromy representation is given by the $T_i$. To address your concerns about this being unique up to isomorphism, notice that a change of basis of $mathcal{O}_X^r$ induces a corresponding (compatible) change to the $T_i$.
You might be interested in the following references: Sections 5.1.1 and 5.1.2 of Hotta, Tanasaki, and Takeuchi's "D-Modules, Perverse Sheaves, and Representation Theory"; and Chapter III of "Algebraic D-Modules" by Borel, et al.
answered Feb 24 at 17:41
Avi SteinerAvi Steiner
1,52211229
$endgroup$
You are correct that the monodromy representation is given by the $T_i$. To address your concerns about this being unique up to isomorphism, notice that a change of basis of $mathcal{O}_X^r$ induces a corresponding (compatible) change to the $T_i$.
You might be interested in the following references: Sections 5.1.1 and 5.1.2 of Hotta, Tanasaki, and Takeuchi's "D-Modules, Perverse Sheaves, and Representation Theory"; and Chapter III of "Algebraic D-Modules" by Borel, et al.
answered Feb 24 at 17:41
Avi SteinerAvi Steiner
1,52211229
answered Feb 24 at 17:41
Avi SteinerAvi Steiner
1,52211229
answered Feb 24 at 17:41
Avi SteinerAvi Steiner
1,52211229
answered Feb 24 at 17:41
Avi SteinerAvi Steiner
1,52211229
1,52211229
$begingroup$
Thanks a lot. Actually the reference you gave is exactly the book I am reading. In this book, they gave a way to obtain the monodromy representation, however, their method requires that all monodromy are commuting invertible matrices, which I do not get the point, do you have any idea?
$endgroup$
– Longma
Feb 25 at 9:03
$begingroup$
@Longma if you don’t have commuting matrices, you don’t get an integrable connection, and in particular can’t use the Riemann-Hilbert Correspondence. In fact, you won’t even get a D-module
$endgroup$
– Avi Steiner
Feb 25 at 16:03
$begingroup$
In higher dimensional cases, I totally agree with you. However, in my example, we are working on the one dimensional case, namely, every two components of the NCD divisor has no intersection, so we don't need to require that the monodromy commutates. (Since the commutativity condition is due to monodromy agrees on the intersection of each "singularity". Does it make any sense for you?
$endgroup$
– Longma
Feb 26 at 11:04
$begingroup$
@Longma Oh! Shoot! You’re right! Yeah, there’s no particular reason you need commuting matrices. My answer still holds , though.
$endgroup$
– Avi Steiner
Feb 26 at 14:21
$begingroup$
Ok. I need more information to determine the induced map between isomorphic classes. Thanks anyway.
$endgroup$
– Longma
Feb 26 at 19:42
add a comment |
$begingroup$
Thanks a lot. Actually the reference you gave is exactly the book I am reading. In this book, they gave a way to obtain the monodromy representation, however, their method requires that all monodromy are commuting invertible matrices, which I do not get the point, do you have any idea?
$endgroup$
– Longma
Feb 25 at 9:03
$begingroup$
@Longma if you don’t have commuting matrices, you don’t get an integrable connection, and in particular can’t use the Riemann-Hilbert Correspondence. In fact, you won’t even get a D-module
$endgroup$
– Avi Steiner
Feb 25 at 16:03
$begingroup$
In higher dimensional cases, I totally agree with you. However, in my example, we are working on the one dimensional case, namely, every two components of the NCD divisor has no intersection, so we don't need to require that the monodromy commutates. (Since the commutativity condition is due to monodromy agrees on the intersection of each "singularity". Does it make any sense for you?
$endgroup$
– Longma
Feb 26 at 11:04
$begingroup$
@Longma Oh! Shoot! You’re right! Yeah, there’s no particular reason you need commuting matrices. My answer still holds , though.
$endgroup$
– Avi Steiner
Feb 26 at 14:21
$begingroup$
Ok. I need more information to determine the induced map between isomorphic classes. Thanks anyway.
$endgroup$
– Longma
Feb 26 at 19:42
$begingroup$
Thanks a lot. Actually the reference you gave is exactly the book I am reading. In this book, they gave a way to obtain the monodromy representation, however, their method requires that all monodromy are commuting invertible matrices, which I do not get the point, do you have any idea?
$endgroup$
– Longma
Feb 25 at 9:03
$begingroup$
Thanks a lot. Actually the reference you gave is exactly the book I am reading. In this book, they gave a way to obtain the monodromy representation, however, their method requires that all monodromy are commuting invertible matrices, which I do not get the point, do you have any idea?
$endgroup$
– Longma
Feb 25 at 9:03
$begingroup$
@Longma if you don’t have commuting matrices, you don’t get an integrable connection, and in particular can’t use the Riemann-Hilbert Correspondence. In fact, you won’t even get a D-module
$endgroup$
– Avi Steiner
Feb 25 at 16:03
$begingroup$
@Longma if you don’t have commuting matrices, you don’t get an integrable connection, and in particular can’t use the Riemann-Hilbert Correspondence. In fact, you won’t even get a D-module
$endgroup$
– Avi Steiner
Feb 25 at 16:03
$begingroup$
In higher dimensional cases, I totally agree with you. However, in my example, we are working on the one dimensional case, namely, every two components of the NCD divisor has no intersection, so we don't need to require that the monodromy commutates. (Since the commutativity condition is due to monodromy agrees on the intersection of each "singularity". Does it make any sense for you?
$endgroup$
– Longma
Feb 26 at 11:04
$begingroup$
In higher dimensional cases, I totally agree with you. However, in my example, we are working on the one dimensional case, namely, every two components of the NCD divisor has no intersection, so we don't need to require that the monodromy commutates. (Since the commutativity condition is due to monodromy agrees on the intersection of each "singularity". Does it make any sense for you?
$endgroup$
– Longma
Feb 26 at 11:04
$begingroup$
@Longma Oh! Shoot! You’re right! Yeah, there’s no particular reason you need commuting matrices. My answer still holds , though.
$endgroup$
– Avi Steiner
Feb 26 at 14:21
$begingroup$
@Longma Oh! Shoot! You’re right! Yeah, there’s no particular reason you need commuting matrices. My answer still holds , though.
$endgroup$
– Avi Steiner
Feb 26 at 14:21
$begingroup$
Ok. I need more information to determine the induced map between isomorphic classes. Thanks anyway.
$endgroup$
– Longma
Feb 26 at 19:42
$begingroup$
Ok. I need more information to determine the induced map between isomorphic classes. Thanks anyway.
$endgroup$
– Longma
Feb 26 at 19:42
add a comment |
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$begingroup$
You cannot solve ode "explicitly" - so cannot find explicit monodromies. Rare exceptions like Knizhnik-Zamolodchikov equation are due some hidden Lie group symmetry inside.
$endgroup$
– Alexander Chervov
Feb 24 at 12:20
$begingroup$
By Riemann Hilbert correspondence, if the sum of A_i is zero, and fix a map $mathbb C/mathbb Zto mathbb C$, then there exists a unique local system, namely a unique representation.
$endgroup$
– Longma
Feb 25 at 9:06