Analytic continuation of several complex variables
$begingroup$
Let $f(w_1,ldots,w_n;z)$ be a holomorphic function of $n+1$ variables.
For every fixed $w_1ldots w_n$, let $g(w_1,ldots,w_n;z)$ be an analytic continuation of $f$ as a holomorphic function of $z$.
Of course, $g$ is holomorphic in $z$.
Now, is $g$ holomorphic in $w_1,ldots, w_n,z$ as well?
In the concrete, I am interested in the following situation.
Let $E(a_1,ldots,a_n;z)$ be an ODE, where $z$ is a complex variable and $a_1,ldots,a_n$ are complex parameters.
Assume that $E(a_1,ldots,a_n;z)$ has a local solution $u(a_1,ldots,a_n;z)$ which is holomorphic in $a_1,ldots,a_n,z$, and $v(a_1,ldots,a_n;z)$ is an analytic continuation of $u$.
Is $v$ holomorphic in $a_1,ldots,a_n,z$?
complex-analysis ordinary-differential-equations several-complex-variables analytic-continuation
$endgroup$
add a comment |
$begingroup$
Let $f(w_1,ldots,w_n;z)$ be a holomorphic function of $n+1$ variables.
For every fixed $w_1ldots w_n$, let $g(w_1,ldots,w_n;z)$ be an analytic continuation of $f$ as a holomorphic function of $z$.
Of course, $g$ is holomorphic in $z$.
Now, is $g$ holomorphic in $w_1,ldots, w_n,z$ as well?
In the concrete, I am interested in the following situation.
Let $E(a_1,ldots,a_n;z)$ be an ODE, where $z$ is a complex variable and $a_1,ldots,a_n$ are complex parameters.
Assume that $E(a_1,ldots,a_n;z)$ has a local solution $u(a_1,ldots,a_n;z)$ which is holomorphic in $a_1,ldots,a_n,z$, and $v(a_1,ldots,a_n;z)$ is an analytic continuation of $u$.
Is $v$ holomorphic in $a_1,ldots,a_n,z$?
complex-analysis ordinary-differential-equations several-complex-variables analytic-continuation
$endgroup$
add a comment |
$begingroup$
Let $f(w_1,ldots,w_n;z)$ be a holomorphic function of $n+1$ variables.
For every fixed $w_1ldots w_n$, let $g(w_1,ldots,w_n;z)$ be an analytic continuation of $f$ as a holomorphic function of $z$.
Of course, $g$ is holomorphic in $z$.
Now, is $g$ holomorphic in $w_1,ldots, w_n,z$ as well?
In the concrete, I am interested in the following situation.
Let $E(a_1,ldots,a_n;z)$ be an ODE, where $z$ is a complex variable and $a_1,ldots,a_n$ are complex parameters.
Assume that $E(a_1,ldots,a_n;z)$ has a local solution $u(a_1,ldots,a_n;z)$ which is holomorphic in $a_1,ldots,a_n,z$, and $v(a_1,ldots,a_n;z)$ is an analytic continuation of $u$.
Is $v$ holomorphic in $a_1,ldots,a_n,z$?
complex-analysis ordinary-differential-equations several-complex-variables analytic-continuation
$endgroup$
Let $f(w_1,ldots,w_n;z)$ be a holomorphic function of $n+1$ variables.
For every fixed $w_1ldots w_n$, let $g(w_1,ldots,w_n;z)$ be an analytic continuation of $f$ as a holomorphic function of $z$.
Of course, $g$ is holomorphic in $z$.
Now, is $g$ holomorphic in $w_1,ldots, w_n,z$ as well?
In the concrete, I am interested in the following situation.
Let $E(a_1,ldots,a_n;z)$ be an ODE, where $z$ is a complex variable and $a_1,ldots,a_n$ are complex parameters.
Assume that $E(a_1,ldots,a_n;z)$ has a local solution $u(a_1,ldots,a_n;z)$ which is holomorphic in $a_1,ldots,a_n,z$, and $v(a_1,ldots,a_n;z)$ is an analytic continuation of $u$.
Is $v$ holomorphic in $a_1,ldots,a_n,z$?
complex-analysis ordinary-differential-equations several-complex-variables analytic-continuation
complex-analysis ordinary-differential-equations several-complex-variables analytic-continuation
edited Dec 18 '18 at 8:26
Daniele Tampieri
2,3272922
2,3272922
asked Aug 12 '18 at 5:31
user356126user356126
1306
1306
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer is yes and it is a consequence of the classical Hartogs‘s theorem on separately holomorphic functions. Precisely, let $u(a_1,dots,a_n,z)$ be a holomorphic function on a domain $D_uinmathbb{C}^{n+1}$ which is analytically continuable to a larger domain respect to the variable $z$ as a function $v(a_1,dots,a_n,z)$: then
$v$ is separately holomorphic respect to $a_1,dots,a_n$ for every $znotin D_u$. To see this, choose $z_0in D_ucap{zinmathbb{C}}$ such that the Taylor series expansion of $v$ in $z_0$ has a convergence disk not entirely contained in $D_ucap{zinmathbb{C}}$ and whose of radius $R_{z_0}geq c>0$ does not depend on $a_1,dots,a_n$. Such $z_0$ exists, since assuming the contrary would deny the possibility of analytically continue $u$ respect to $z$ outside $D_ucap{zinmathbb{C}} $.
Then, evaluating this Taylor series at fixed
point $z_1notin D_ucap{zinmathbb{C}} $ inside its radius of convergence, we get
$$
begin{split}
v(a_1,dots,a_n, z_1)&=sum_{k=0}^inftyfrac{1}{k!} frac{partial^k v}{partial z^k}(a_1,dots,a_n, z_0)(z_1-z_0)^k\
&=sum_{k=0}^inftyfrac{1}{k!} frac{partial^k u}{partial z^k}(a_1,dots,a_n, z_0)(z_1-z_0)^k
end{split}tag{1}label{1}
$$
since $v=u$ on $D_u$. Now eqref{1} implies that the $N$-th order Taylor polynomial
$$
v_N(dots,a_j,dots,z_1)=sum_{k=0}^Nfrac{1}{k!} frac{partial^k u}{partial z^k}(dots,a_j,dots, z_0)(z_1-z_0)^ktag{2}label{2}
$$
can be considered as a sequence of holomorphic functions in each single variable $a_j$, converging uniformly (thanks to the holomophicity of $u$ and to the uniform convergence of eqref{1}) to $v$. This suffices to prove the separate analyticity of $v$, analytic continuation of $u$, respect to all its variables in the domain
$$
D_ucup{(a_1,dots,a_n,z)|0<|z-z_0|leq |z_1-z_0|}.
$$
The same argument can be repeated to cover all the domain $D_v$, as quckly sketched in the following picture:
- $v$ is separately holomorphic respect to each of the variables $a_1,dots,a_n, zin D_vvarsupsetneq D_u$ thus, by Hartogs’s theorem, it is jointly holomorphic respect to all its variables on $D_v$.
$endgroup$
$begingroup$
Thank you for your answer. I didn't hit upon the idea of using Taylor series.
$endgroup$
– user356126
Aug 17 '18 at 10:03
$begingroup$
You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
$endgroup$
– Daniele Tampieri
Aug 17 '18 at 14:56
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2880002%2fanalytic-continuation-of-several-complex-variables%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is yes and it is a consequence of the classical Hartogs‘s theorem on separately holomorphic functions. Precisely, let $u(a_1,dots,a_n,z)$ be a holomorphic function on a domain $D_uinmathbb{C}^{n+1}$ which is analytically continuable to a larger domain respect to the variable $z$ as a function $v(a_1,dots,a_n,z)$: then
$v$ is separately holomorphic respect to $a_1,dots,a_n$ for every $znotin D_u$. To see this, choose $z_0in D_ucap{zinmathbb{C}}$ such that the Taylor series expansion of $v$ in $z_0$ has a convergence disk not entirely contained in $D_ucap{zinmathbb{C}}$ and whose of radius $R_{z_0}geq c>0$ does not depend on $a_1,dots,a_n$. Such $z_0$ exists, since assuming the contrary would deny the possibility of analytically continue $u$ respect to $z$ outside $D_ucap{zinmathbb{C}} $.
Then, evaluating this Taylor series at fixed
point $z_1notin D_ucap{zinmathbb{C}} $ inside its radius of convergence, we get
$$
begin{split}
v(a_1,dots,a_n, z_1)&=sum_{k=0}^inftyfrac{1}{k!} frac{partial^k v}{partial z^k}(a_1,dots,a_n, z_0)(z_1-z_0)^k\
&=sum_{k=0}^inftyfrac{1}{k!} frac{partial^k u}{partial z^k}(a_1,dots,a_n, z_0)(z_1-z_0)^k
end{split}tag{1}label{1}
$$
since $v=u$ on $D_u$. Now eqref{1} implies that the $N$-th order Taylor polynomial
$$
v_N(dots,a_j,dots,z_1)=sum_{k=0}^Nfrac{1}{k!} frac{partial^k u}{partial z^k}(dots,a_j,dots, z_0)(z_1-z_0)^ktag{2}label{2}
$$
can be considered as a sequence of holomorphic functions in each single variable $a_j$, converging uniformly (thanks to the holomophicity of $u$ and to the uniform convergence of eqref{1}) to $v$. This suffices to prove the separate analyticity of $v$, analytic continuation of $u$, respect to all its variables in the domain
$$
D_ucup{(a_1,dots,a_n,z)|0<|z-z_0|leq |z_1-z_0|}.
$$
The same argument can be repeated to cover all the domain $D_v$, as quckly sketched in the following picture:
- $v$ is separately holomorphic respect to each of the variables $a_1,dots,a_n, zin D_vvarsupsetneq D_u$ thus, by Hartogs’s theorem, it is jointly holomorphic respect to all its variables on $D_v$.
$endgroup$
$begingroup$
Thank you for your answer. I didn't hit upon the idea of using Taylor series.
$endgroup$
– user356126
Aug 17 '18 at 10:03
$begingroup$
You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
$endgroup$
– Daniele Tampieri
Aug 17 '18 at 14:56
add a comment |
$begingroup$
The answer is yes and it is a consequence of the classical Hartogs‘s theorem on separately holomorphic functions. Precisely, let $u(a_1,dots,a_n,z)$ be a holomorphic function on a domain $D_uinmathbb{C}^{n+1}$ which is analytically continuable to a larger domain respect to the variable $z$ as a function $v(a_1,dots,a_n,z)$: then
$v$ is separately holomorphic respect to $a_1,dots,a_n$ for every $znotin D_u$. To see this, choose $z_0in D_ucap{zinmathbb{C}}$ such that the Taylor series expansion of $v$ in $z_0$ has a convergence disk not entirely contained in $D_ucap{zinmathbb{C}}$ and whose of radius $R_{z_0}geq c>0$ does not depend on $a_1,dots,a_n$. Such $z_0$ exists, since assuming the contrary would deny the possibility of analytically continue $u$ respect to $z$ outside $D_ucap{zinmathbb{C}} $.
Then, evaluating this Taylor series at fixed
point $z_1notin D_ucap{zinmathbb{C}} $ inside its radius of convergence, we get
$$
begin{split}
v(a_1,dots,a_n, z_1)&=sum_{k=0}^inftyfrac{1}{k!} frac{partial^k v}{partial z^k}(a_1,dots,a_n, z_0)(z_1-z_0)^k\
&=sum_{k=0}^inftyfrac{1}{k!} frac{partial^k u}{partial z^k}(a_1,dots,a_n, z_0)(z_1-z_0)^k
end{split}tag{1}label{1}
$$
since $v=u$ on $D_u$. Now eqref{1} implies that the $N$-th order Taylor polynomial
$$
v_N(dots,a_j,dots,z_1)=sum_{k=0}^Nfrac{1}{k!} frac{partial^k u}{partial z^k}(dots,a_j,dots, z_0)(z_1-z_0)^ktag{2}label{2}
$$
can be considered as a sequence of holomorphic functions in each single variable $a_j$, converging uniformly (thanks to the holomophicity of $u$ and to the uniform convergence of eqref{1}) to $v$. This suffices to prove the separate analyticity of $v$, analytic continuation of $u$, respect to all its variables in the domain
$$
D_ucup{(a_1,dots,a_n,z)|0<|z-z_0|leq |z_1-z_0|}.
$$
The same argument can be repeated to cover all the domain $D_v$, as quckly sketched in the following picture:
- $v$ is separately holomorphic respect to each of the variables $a_1,dots,a_n, zin D_vvarsupsetneq D_u$ thus, by Hartogs’s theorem, it is jointly holomorphic respect to all its variables on $D_v$.
$endgroup$
$begingroup$
Thank you for your answer. I didn't hit upon the idea of using Taylor series.
$endgroup$
– user356126
Aug 17 '18 at 10:03
$begingroup$
You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
$endgroup$
– Daniele Tampieri
Aug 17 '18 at 14:56
add a comment |
$begingroup$
The answer is yes and it is a consequence of the classical Hartogs‘s theorem on separately holomorphic functions. Precisely, let $u(a_1,dots,a_n,z)$ be a holomorphic function on a domain $D_uinmathbb{C}^{n+1}$ which is analytically continuable to a larger domain respect to the variable $z$ as a function $v(a_1,dots,a_n,z)$: then
$v$ is separately holomorphic respect to $a_1,dots,a_n$ for every $znotin D_u$. To see this, choose $z_0in D_ucap{zinmathbb{C}}$ such that the Taylor series expansion of $v$ in $z_0$ has a convergence disk not entirely contained in $D_ucap{zinmathbb{C}}$ and whose of radius $R_{z_0}geq c>0$ does not depend on $a_1,dots,a_n$. Such $z_0$ exists, since assuming the contrary would deny the possibility of analytically continue $u$ respect to $z$ outside $D_ucap{zinmathbb{C}} $.
Then, evaluating this Taylor series at fixed
point $z_1notin D_ucap{zinmathbb{C}} $ inside its radius of convergence, we get
$$
begin{split}
v(a_1,dots,a_n, z_1)&=sum_{k=0}^inftyfrac{1}{k!} frac{partial^k v}{partial z^k}(a_1,dots,a_n, z_0)(z_1-z_0)^k\
&=sum_{k=0}^inftyfrac{1}{k!} frac{partial^k u}{partial z^k}(a_1,dots,a_n, z_0)(z_1-z_0)^k
end{split}tag{1}label{1}
$$
since $v=u$ on $D_u$. Now eqref{1} implies that the $N$-th order Taylor polynomial
$$
v_N(dots,a_j,dots,z_1)=sum_{k=0}^Nfrac{1}{k!} frac{partial^k u}{partial z^k}(dots,a_j,dots, z_0)(z_1-z_0)^ktag{2}label{2}
$$
can be considered as a sequence of holomorphic functions in each single variable $a_j$, converging uniformly (thanks to the holomophicity of $u$ and to the uniform convergence of eqref{1}) to $v$. This suffices to prove the separate analyticity of $v$, analytic continuation of $u$, respect to all its variables in the domain
$$
D_ucup{(a_1,dots,a_n,z)|0<|z-z_0|leq |z_1-z_0|}.
$$
The same argument can be repeated to cover all the domain $D_v$, as quckly sketched in the following picture:
- $v$ is separately holomorphic respect to each of the variables $a_1,dots,a_n, zin D_vvarsupsetneq D_u$ thus, by Hartogs’s theorem, it is jointly holomorphic respect to all its variables on $D_v$.
$endgroup$
The answer is yes and it is a consequence of the classical Hartogs‘s theorem on separately holomorphic functions. Precisely, let $u(a_1,dots,a_n,z)$ be a holomorphic function on a domain $D_uinmathbb{C}^{n+1}$ which is analytically continuable to a larger domain respect to the variable $z$ as a function $v(a_1,dots,a_n,z)$: then
$v$ is separately holomorphic respect to $a_1,dots,a_n$ for every $znotin D_u$. To see this, choose $z_0in D_ucap{zinmathbb{C}}$ such that the Taylor series expansion of $v$ in $z_0$ has a convergence disk not entirely contained in $D_ucap{zinmathbb{C}}$ and whose of radius $R_{z_0}geq c>0$ does not depend on $a_1,dots,a_n$. Such $z_0$ exists, since assuming the contrary would deny the possibility of analytically continue $u$ respect to $z$ outside $D_ucap{zinmathbb{C}} $.
Then, evaluating this Taylor series at fixed
point $z_1notin D_ucap{zinmathbb{C}} $ inside its radius of convergence, we get
$$
begin{split}
v(a_1,dots,a_n, z_1)&=sum_{k=0}^inftyfrac{1}{k!} frac{partial^k v}{partial z^k}(a_1,dots,a_n, z_0)(z_1-z_0)^k\
&=sum_{k=0}^inftyfrac{1}{k!} frac{partial^k u}{partial z^k}(a_1,dots,a_n, z_0)(z_1-z_0)^k
end{split}tag{1}label{1}
$$
since $v=u$ on $D_u$. Now eqref{1} implies that the $N$-th order Taylor polynomial
$$
v_N(dots,a_j,dots,z_1)=sum_{k=0}^Nfrac{1}{k!} frac{partial^k u}{partial z^k}(dots,a_j,dots, z_0)(z_1-z_0)^ktag{2}label{2}
$$
can be considered as a sequence of holomorphic functions in each single variable $a_j$, converging uniformly (thanks to the holomophicity of $u$ and to the uniform convergence of eqref{1}) to $v$. This suffices to prove the separate analyticity of $v$, analytic continuation of $u$, respect to all its variables in the domain
$$
D_ucup{(a_1,dots,a_n,z)|0<|z-z_0|leq |z_1-z_0|}.
$$
The same argument can be repeated to cover all the domain $D_v$, as quckly sketched in the following picture:
- $v$ is separately holomorphic respect to each of the variables $a_1,dots,a_n, zin D_vvarsupsetneq D_u$ thus, by Hartogs’s theorem, it is jointly holomorphic respect to all its variables on $D_v$.
edited Aug 14 '18 at 9:42
answered Aug 12 '18 at 20:06
Daniele TampieriDaniele Tampieri
2,3272922
2,3272922
$begingroup$
Thank you for your answer. I didn't hit upon the idea of using Taylor series.
$endgroup$
– user356126
Aug 17 '18 at 10:03
$begingroup$
You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
$endgroup$
– Daniele Tampieri
Aug 17 '18 at 14:56
add a comment |
$begingroup$
Thank you for your answer. I didn't hit upon the idea of using Taylor series.
$endgroup$
– user356126
Aug 17 '18 at 10:03
$begingroup$
You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
$endgroup$
– Daniele Tampieri
Aug 17 '18 at 14:56
$begingroup$
Thank you for your answer. I didn't hit upon the idea of using Taylor series.
$endgroup$
– user356126
Aug 17 '18 at 10:03
$begingroup$
Thank you for your answer. I didn't hit upon the idea of using Taylor series.
$endgroup$
– user356126
Aug 17 '18 at 10:03
$begingroup$
You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
$endgroup$
– Daniele Tampieri
Aug 17 '18 at 14:56
$begingroup$
You're welcome. Using this kind of argument is a standard way of proving results on analytic continuation in the elementary theory of holomorphic functions of several variables: and as such you probably already noticed its main problem. We have almost no information on the size of $D_v$, i.e. the analytic continuation results obtained in this way do not say much on the structure of the domain where the function can be continued.
$endgroup$
– Daniele Tampieri
Aug 17 '18 at 14:56
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2880002%2fanalytic-continuation-of-several-complex-variables%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown