Sheaf of Meromorphic Functions
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In one of the exercise sheets for my complex analysis course we are given the following task
Prove that the set of all meromorphic functions on $mathbb{C}$ defines a (pre)sheaf.
showing that they form a presheaf is fairly trivial. In order to check the second sheaf Axiom (the gluing Axiom - at least Wikipedia calls is so) the solution suggests that one has to use Mittag-Leffler and the identity theorem.
However I think I can provide an "Elementary" proof: Let $mathcal{M}$ denote the Sheaf of meromorphic functions, $U subset mathbb{C}$ be open and ${U_i}_{i in I}$ be an open cover of $U$.
I have to show that if a familiy $f_i in mathcal{M}(U_i)$ satisfies $f_i|_{U_i cap U_j} = f_j|_{U_i cap U_j}$ for all $i, j in I$ then there exists an $f in mathcal{M}(U)$ such that $forall i in I: f|_{U_i} = f_i$.
Now as fas as I can see I can construct a function $f: U to mathbb{C} cup {infty}$ which maps any $x in U$ to an $f_i(x)$ where $i$ is chosen such that $x in U_i$. This function satisfies the aforementioned property so we only have to check that it is in fact meromorphic. For every $x$ there is an $U_i$ which contains $x$ and therefore also a neighborhood $V$ of $x$. Since $f_i|_V$ is meromorphic it follows, that $f|_V = f_i|_V$ is meromorphic. Since meromorphy is a local property $f$ as a whole is meromorphic.
Is this correct?
I am not at all familiar with Sheafs so please point out if there is any wrong use of notation in my question.
complex-analysis proof-verification sheaf-theory meromorphic-functions
asked Mar 8 '18 at 21:33
0x5390x539
1,361518
$endgroup$
add a comment |
$begingroup$
In one of the exercise sheets for my complex analysis course we are given the following task
Prove that the set of all meromorphic functions on $mathbb{C}$ defines a (pre)sheaf.
showing that they form a presheaf is fairly trivial. In order to check the second sheaf Axiom (the gluing Axiom - at least Wikipedia calls is so) the solution suggests that one has to use Mittag-Leffler and the identity theorem.
However I think I can provide an "Elementary" proof: Let $mathcal{M}$ denote the Sheaf of meromorphic functions, $U subset mathbb{C}$ be open and ${U_i}_{i in I}$ be an open cover of $U$.
I have to show that if a familiy $f_i in mathcal{M}(U_i)$ satisfies $f_i|_{U_i cap U_j} = f_j|_{U_i cap U_j}$ for all $i, j in I$ then there exists an $f in mathcal{M}(U)$ such that $forall i in I: f|_{U_i} = f_i$.
Now as fas as I can see I can construct a function $f: U to mathbb{C} cup {infty}$ which maps any $x in U$ to an $f_i(x)$ where $i$ is chosen such that $x in U_i$. This function satisfies the aforementioned property so we only have to check that it is in fact meromorphic. For every $x$ there is an $U_i$ which contains $x$ and therefore also a neighborhood $V$ of $x$. Since $f_i|_V$ is meromorphic it follows, that $f|_V = f_i|_V$ is meromorphic. Since meromorphy is a local property $f$ as a whole is meromorphic.
Is this correct?
I am not at all familiar with Sheafs so please point out if there is any wrong use of notation in my question.
complex-analysis proof-verification sheaf-theory meromorphic-functions
asked Mar 8 '18 at 21:33
0x5390x539
1,361518
$endgroup$
$begingroup$
In order to prove that it is a sheaf, you also need to show that this $f$ is unique, hence you need to use the identity theorem.
$endgroup$
– Levent
Mar 8 '18 at 21:51
$begingroup$
@Levent AFAIK that follows from the first sheaf axiom
$endgroup$
– 0x539
Mar 8 '18 at 22:11
$begingroup$
Yes, your elementary proof is correct and the reference to Mittag-Leffler is indeed absurdly irrelevant. The point I wanted to emphasize in my erased comments is that $mathcal M(U)$ consists of holomorphic maps $Uto mathbb Ccup {infty}$ which take the value $infty$ only on a discrete closed subset (maybe empty) of $U$. A typical example of meromorphic function in $mathcal M (mathbb C) $ is the function $frac {1}{sin z}$, for which the discrete subset is $pi mathbb Z$.
$endgroup$
– Georges Elencwajg
Mar 8 '18 at 23:36
add a comment |
$begingroup$
In one of the exercise sheets for my complex analysis course we are given the following task
Prove that the set of all meromorphic functions on $mathbb{C}$ defines a (pre)sheaf.
showing that they form a presheaf is fairly trivial. In order to check the second sheaf Axiom (the gluing Axiom - at least Wikipedia calls is so) the solution suggests that one has to use Mittag-Leffler and the identity theorem.
However I think I can provide an "Elementary" proof: Let $mathcal{M}$ denote the Sheaf of meromorphic functions, $U subset mathbb{C}$ be open and ${U_i}_{i in I}$ be an open cover of $U$.
I have to show that if a familiy $f_i in mathcal{M}(U_i)$ satisfies $f_i|_{U_i cap U_j} = f_j|_{U_i cap U_j}$ for all $i, j in I$ then there exists an $f in mathcal{M}(U)$ such that $forall i in I: f|_{U_i} = f_i$.
Now as fas as I can see I can construct a function $f: U to mathbb{C} cup {infty}$ which maps any $x in U$ to an $f_i(x)$ where $i$ is chosen such that $x in U_i$. This function satisfies the aforementioned property so we only have to check that it is in fact meromorphic. For every $x$ there is an $U_i$ which contains $x$ and therefore also a neighborhood $V$ of $x$. Since $f_i|_V$ is meromorphic it follows, that $f|_V = f_i|_V$ is meromorphic. Since meromorphy is a local property $f$ as a whole is meromorphic.
Is this correct?
I am not at all familiar with Sheafs so please point out if there is any wrong use of notation in my question.
complex-analysis proof-verification sheaf-theory meromorphic-functions
asked Mar 8 '18 at 21:33
0x5390x539
1,361518
$endgroup$
In one of the exercise sheets for my complex analysis course we are given the following task
Prove that the set of all meromorphic functions on $mathbb{C}$ defines a (pre)sheaf.
showing that they form a presheaf is fairly trivial. In order to check the second sheaf Axiom (the gluing Axiom - at least Wikipedia calls is so) the solution suggests that one has to use Mittag-Leffler and the identity theorem.
However I think I can provide an "Elementary" proof: Let $mathcal{M}$ denote the Sheaf of meromorphic functions, $U subset mathbb{C}$ be open and ${U_i}_{i in I}$ be an open cover of $U$.
I have to show that if a familiy $f_i in mathcal{M}(U_i)$ satisfies $f_i|_{U_i cap U_j} = f_j|_{U_i cap U_j}$ for all $i, j in I$ then there exists an $f in mathcal{M}(U)$ such that $forall i in I: f|_{U_i} = f_i$.
Now as fas as I can see I can construct a function $f: U to mathbb{C} cup {infty}$ which maps any $x in U$ to an $f_i(x)$ where $i$ is chosen such that $x in U_i$. This function satisfies the aforementioned property so we only have to check that it is in fact meromorphic. For every $x$ there is an $U_i$ which contains $x$ and therefore also a neighborhood $V$ of $x$. Since $f_i|_V$ is meromorphic it follows, that $f|_V = f_i|_V$ is meromorphic. Since meromorphy is a local property $f$ as a whole is meromorphic.
Is this correct?
I am not at all familiar with Sheafs so please point out if there is any wrong use of notation in my question.
complex-analysis proof-verification sheaf-theory meromorphic-functions
complex-analysis proof-verification sheaf-theory meromorphic-functions
asked Mar 8 '18 at 21:33
0x5390x539
1,361518
asked Mar 8 '18 at 21:33
0x5390x539
1,361518
edited Dec 8 '18 at 15:04
0x539
asked Mar 8 '18 at 21:33
0x5390x539
1,361518
asked Mar 8 '18 at 21:33
0x5390x539
1,361518
asked Mar 8 '18 at 21:33
0x5390x539
1,361518
1,361518
$begingroup$
In order to prove that it is a sheaf, you also need to show that this $f$ is unique, hence you need to use the identity theorem.
$endgroup$
– Levent
Mar 8 '18 at 21:51
$begingroup$
@Levent AFAIK that follows from the first sheaf axiom
$endgroup$
– 0x539
Mar 8 '18 at 22:11
$begingroup$
Yes, your elementary proof is correct and the reference to Mittag-Leffler is indeed absurdly irrelevant. The point I wanted to emphasize in my erased comments is that $mathcal M(U)$ consists of holomorphic maps $Uto mathbb Ccup {infty}$ which take the value $infty$ only on a discrete closed subset (maybe empty) of $U$. A typical example of meromorphic function in $mathcal M (mathbb C) $ is the function $frac {1}{sin z}$, for which the discrete subset is $pi mathbb Z$.
$endgroup$
– Georges Elencwajg
Mar 8 '18 at 23:36
add a comment |
$begingroup$
In order to prove that it is a sheaf, you also need to show that this $f$ is unique, hence you need to use the identity theorem.
$endgroup$
– Levent
Mar 8 '18 at 21:51
$begingroup$
@Levent AFAIK that follows from the first sheaf axiom
$endgroup$
– 0x539
Mar 8 '18 at 22:11
$begingroup$
Yes, your elementary proof is correct and the reference to Mittag-Leffler is indeed absurdly irrelevant. The point I wanted to emphasize in my erased comments is that $mathcal M(U)$ consists of holomorphic maps $Uto mathbb Ccup {infty}$ which take the value $infty$ only on a discrete closed subset (maybe empty) of $U$. A typical example of meromorphic function in $mathcal M (mathbb C) $ is the function $frac {1}{sin z}$, for which the discrete subset is $pi mathbb Z$.
$endgroup$
– Georges Elencwajg
Mar 8 '18 at 23:36
$begingroup$
In order to prove that it is a sheaf, you also need to show that this $f$ is unique, hence you need to use the identity theorem.
$endgroup$
– Levent
Mar 8 '18 at 21:51
$begingroup$
In order to prove that it is a sheaf, you also need to show that this $f$ is unique, hence you need to use the identity theorem.
$endgroup$
– Levent
Mar 8 '18 at 21:51
$begingroup$
@Levent AFAIK that follows from the first sheaf axiom
$endgroup$
– 0x539
Mar 8 '18 at 22:11
$begingroup$
@Levent AFAIK that follows from the first sheaf axiom
$endgroup$
– 0x539
Mar 8 '18 at 22:11
$begingroup$
Yes, your elementary proof is correct and the reference to Mittag-Leffler is indeed absurdly irrelevant. The point I wanted to emphasize in my erased comments is that $mathcal M(U)$ consists of holomorphic maps $Uto mathbb Ccup {infty}$ which take the value $infty$ only on a discrete closed subset (maybe empty) of $U$. A typical example of meromorphic function in $mathcal M (mathbb C) $ is the function $frac {1}{sin z}$, for which the discrete subset is $pi mathbb Z$.
$endgroup$
– Georges Elencwajg
Mar 8 '18 at 23:36
$begingroup$
Yes, your elementary proof is correct and the reference to Mittag-Leffler is indeed absurdly irrelevant. The point I wanted to emphasize in my erased comments is that $mathcal M(U)$ consists of holomorphic maps $Uto mathbb Ccup {infty}$ which take the value $infty$ only on a discrete closed subset (maybe empty) of $U$. A typical example of meromorphic function in $mathcal M (mathbb C) $ is the function $frac {1}{sin z}$, for which the discrete subset is $pi mathbb Z$.
$endgroup$
– Georges Elencwajg
Mar 8 '18 at 23:36
add a comment |
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$begingroup$
In order to prove that it is a sheaf, you also need to show that this $f$ is unique, hence you need to use the identity theorem.
$endgroup$
– Levent
Mar 8 '18 at 21:51
$begingroup$
@Levent AFAIK that follows from the first sheaf axiom
$endgroup$
– 0x539
Mar 8 '18 at 22:11
$begingroup$
Yes, your elementary proof is correct and the reference to Mittag-Leffler is indeed absurdly irrelevant. The point I wanted to emphasize in my erased comments is that $mathcal M(U)$ consists of holomorphic maps $Uto mathbb Ccup {infty}$ which take the value $infty$ only on a discrete closed subset (maybe empty) of $U$. A typical example of meromorphic function in $mathcal M (mathbb C) $ is the function $frac {1}{sin z}$, for which the discrete subset is $pi mathbb Z$.
$endgroup$
– Georges Elencwajg
Mar 8 '18 at 23:36