Mixing
Number of minima in a ribbon disk?
$begingroup$
I am asking this question mainly with the hope of finding a reference to (presumably well-trodden) topic. Let $K$ be a ribbon knot and define $I(K)$ to be the minimum over number of minima of all ribbon disks of $K$ ("I" for invariant - I'm guessing that it has a name and a literature).
Is there a family of ribbon knots that show that $I(K)$ can be arbitrarily large? In other words, how do I get a hold on lower bounds for $I(K)$ - hopefully without assuming slice = ribbon.
geometric-topology knot-theory knot-invariants
asked Jan 27 at 23:28
user101010user101010
1,953416
$endgroup$
add a comment |
$begingroup$
I am asking this question mainly with the hope of finding a reference to (presumably well-trodden) topic. Let $K$ be a ribbon knot and define $I(K)$ to be the minimum over number of minima of all ribbon disks of $K$ ("I" for invariant - I'm guessing that it has a name and a literature).
Is there a family of ribbon knots that show that $I(K)$ can be arbitrarily large? In other words, how do I get a hold on lower bounds for $I(K)$ - hopefully without assuming slice = ribbon.
geometric-topology knot-theory knot-invariants
asked Jan 27 at 23:28
user101010user101010
1,953416
$endgroup$
add a comment |
$begingroup$
I am asking this question mainly with the hope of finding a reference to (presumably well-trodden) topic. Let $K$ be a ribbon knot and define $I(K)$ to be the minimum over number of minima of all ribbon disks of $K$ ("I" for invariant - I'm guessing that it has a name and a literature).
Is there a family of ribbon knots that show that $I(K)$ can be arbitrarily large? In other words, how do I get a hold on lower bounds for $I(K)$ - hopefully without assuming slice = ribbon.
geometric-topology knot-theory knot-invariants
asked Jan 27 at 23:28
user101010user101010
1,953416
$endgroup$
I am asking this question mainly with the hope of finding a reference to (presumably well-trodden) topic. Let $K$ be a ribbon knot and define $I(K)$ to be the minimum over number of minima of all ribbon disks of $K$ ("I" for invariant - I'm guessing that it has a name and a literature).
Is there a family of ribbon knots that show that $I(K)$ can be arbitrarily large? In other words, how do I get a hold on lower bounds for $I(K)$ - hopefully without assuming slice = ribbon.
geometric-topology knot-theory knot-invariants
geometric-topology knot-theory knot-invariants
asked Jan 27 at 23:28
user101010user101010
1,953416
asked Jan 27 at 23:28
user101010user101010
1,953416
asked Jan 27 at 23:28
user101010user101010
1,953416
asked Jan 27 at 23:28
user101010user101010
1,953416
asked Jan 27 at 23:28
user101010user101010
1,953416
1,953416
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A handle decomposition of a ribbon disk gives a band presentation of the knot. If I understand how this goes correctly, the band presentation is an unlink along with attached bands between them such that
performing embedded arc surgery along these bands (which carry framing information for the surgery) gives the knot, and
if the unknots are thought of as vertices and the bands as edges, the graph forms a tree (since loops in the graph would correspond to a maxima, which don't exist).
The number of components in the unlink is the number of minima, and so your $I(K)$ is the minimal number of such components over all band presentations of the ribbon knot $K$.
For example, here is a band presentation of a square knot:
By the way, there appear to be some moves that can go between any two band presentations for the same ribbon disk of a ribbon knot:
(See at least Figure 3 of https://arxiv.org/abs/1804.09169 or the paper they cite:
Swenton, Frank J., On a calculus for 2-knots and surfaces in 4-space, J. Knot Theory Ramifications 10, No. 8, 1133-1141 (2001). ZBL1001.57044.). I wonder what additional moves you might need to go between any two ribbon disks.
Anyway, there is a nice way to calculate a group that is the homomorphic image of the fundamental group of the knot complement from such a band presentation, as you are probably aware:
Johnson, Dennis, Homomorphs of knot groups, Proc. Am. Math. Soc. 78, 135-138 (1980). ZBL0435.57003.
Each unknot is assigned a generator, and each band sum introduces a single additional relation depending on which unknots the band passes through. This implies that $I(K)geq operatorname{rank}(pi_1(S^3-K))$, where the rank is the minimal number of generators over all presentations of the finitely generated group $pi_1(S^3-K)$.
The following paper implies that there are knot groups of arbitrary rank:
Weidmann, Richard, On the rank of amalgamated products and product knot groups, Math. Ann. 312, No. 4, 761-771 (1998). ZBL0926.20019.
In particular, the $n$-fold connect sum of any nontrivial ribbon knot $K$ has $I(mathop{#}_nK)geq operatorname{rank}(pi_1(S^3-mathop{#}_nK))geq n+1$.
(If there were a way to modify Johnson's construction to generate prime knots, then one could get an infinite family of prime ribbon knots with arbitrary rank.)
answered Feb 4 at 1:48
Kyle MillerKyle Miller
9,700930
$endgroup$
1
$begingroup$
Thanks Kyle, this is great! I had forgotten about the Johnson paper and I did not know about that result of Weidmann - that is exactly the sort of fact that I was looking for!!
$endgroup$
– user101010
Feb 5 at 0:29
$begingroup$
@user101010 I've realized I misunderstood the kind of group presentation you get from Johnson's construction and that you only get particular quotients of the fundamental group; the ranks of these groups are lower bounds for the rank of the fundamental group itself. I think $I(mathop{#}_n K)geq n+1$ still applies, though.
$endgroup$
– Kyle Miller
Feb 7 at 0:05
add a comment |
$begingroup$
Since the Johnson paper only gives a quotient of the knot group, it is not clear why $I(#_n K)geq n$, and in particular, it is not clear if $I$ can be arbitrarily large. Am I missing something obvious?
answered Mar 22 at 21:16
Sucharit SarkarSucharit Sarkar
101
New contributor
Sucharit Sarkar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
A handle decomposition of a ribbon disk gives a band presentation of the knot. If I understand how this goes correctly, the band presentation is an unlink along with attached bands between them such that
performing embedded arc surgery along these bands (which carry framing information for the surgery) gives the knot, and
if the unknots are thought of as vertices and the bands as edges, the graph forms a tree (since loops in the graph would correspond to a maxima, which don't exist).
The number of components in the unlink is the number of minima, and so your $I(K)$ is the minimal number of such components over all band presentations of the ribbon knot $K$.
For example, here is a band presentation of a square knot:
By the way, there appear to be some moves that can go between any two band presentations for the same ribbon disk of a ribbon knot:
(See at least Figure 3 of https://arxiv.org/abs/1804.09169 or the paper they cite:
Swenton, Frank J., On a calculus for 2-knots and surfaces in 4-space, J. Knot Theory Ramifications 10, No. 8, 1133-1141 (2001). ZBL1001.57044.). I wonder what additional moves you might need to go between any two ribbon disks.
Anyway, there is a nice way to calculate a group that is the homomorphic image of the fundamental group of the knot complement from such a band presentation, as you are probably aware:
Johnson, Dennis, Homomorphs of knot groups, Proc. Am. Math. Soc. 78, 135-138 (1980). ZBL0435.57003.
Each unknot is assigned a generator, and each band sum introduces a single additional relation depending on which unknots the band passes through. This implies that $I(K)geq operatorname{rank}(pi_1(S^3-K))$, where the rank is the minimal number of generators over all presentations of the finitely generated group $pi_1(S^3-K)$.
The following paper implies that there are knot groups of arbitrary rank:
Weidmann, Richard, On the rank of amalgamated products and product knot groups, Math. Ann. 312, No. 4, 761-771 (1998). ZBL0926.20019.
In particular, the $n$-fold connect sum of any nontrivial ribbon knot $K$ has $I(mathop{#}_nK)geq operatorname{rank}(pi_1(S^3-mathop{#}_nK))geq n+1$.
(If there were a way to modify Johnson's construction to generate prime knots, then one could get an infinite family of prime ribbon knots with arbitrary rank.)
answered Feb 4 at 1:48
Kyle MillerKyle Miller
9,700930
$endgroup$
1
$begingroup$
Thanks Kyle, this is great! I had forgotten about the Johnson paper and I did not know about that result of Weidmann - that is exactly the sort of fact that I was looking for!!
$endgroup$
– user101010
Feb 5 at 0:29
$begingroup$
@user101010 I've realized I misunderstood the kind of group presentation you get from Johnson's construction and that you only get particular quotients of the fundamental group; the ranks of these groups are lower bounds for the rank of the fundamental group itself. I think $I(mathop{#}_n K)geq n+1$ still applies, though.
$endgroup$
– Kyle Miller
Feb 7 at 0:05
add a comment |
$begingroup$
A handle decomposition of a ribbon disk gives a band presentation of the knot. If I understand how this goes correctly, the band presentation is an unlink along with attached bands between them such that
performing embedded arc surgery along these bands (which carry framing information for the surgery) gives the knot, and
if the unknots are thought of as vertices and the bands as edges, the graph forms a tree (since loops in the graph would correspond to a maxima, which don't exist).
The number of components in the unlink is the number of minima, and so your $I(K)$ is the minimal number of such components over all band presentations of the ribbon knot $K$.
For example, here is a band presentation of a square knot:
By the way, there appear to be some moves that can go between any two band presentations for the same ribbon disk of a ribbon knot:
(See at least Figure 3 of https://arxiv.org/abs/1804.09169 or the paper they cite:
Swenton, Frank J., On a calculus for 2-knots and surfaces in 4-space, J. Knot Theory Ramifications 10, No. 8, 1133-1141 (2001). ZBL1001.57044.). I wonder what additional moves you might need to go between any two ribbon disks.
Anyway, there is a nice way to calculate a group that is the homomorphic image of the fundamental group of the knot complement from such a band presentation, as you are probably aware:
Johnson, Dennis, Homomorphs of knot groups, Proc. Am. Math. Soc. 78, 135-138 (1980). ZBL0435.57003.
Each unknot is assigned a generator, and each band sum introduces a single additional relation depending on which unknots the band passes through. This implies that $I(K)geq operatorname{rank}(pi_1(S^3-K))$, where the rank is the minimal number of generators over all presentations of the finitely generated group $pi_1(S^3-K)$.
The following paper implies that there are knot groups of arbitrary rank:
Weidmann, Richard, On the rank of amalgamated products and product knot groups, Math. Ann. 312, No. 4, 761-771 (1998). ZBL0926.20019.
In particular, the $n$-fold connect sum of any nontrivial ribbon knot $K$ has $I(mathop{#}_nK)geq operatorname{rank}(pi_1(S^3-mathop{#}_nK))geq n+1$.
(If there were a way to modify Johnson's construction to generate prime knots, then one could get an infinite family of prime ribbon knots with arbitrary rank.)
answered Feb 4 at 1:48
Kyle MillerKyle Miller
9,700930
$endgroup$
1
$begingroup$
Thanks Kyle, this is great! I had forgotten about the Johnson paper and I did not know about that result of Weidmann - that is exactly the sort of fact that I was looking for!!
$endgroup$
– user101010
Feb 5 at 0:29
$begingroup$
@user101010 I've realized I misunderstood the kind of group presentation you get from Johnson's construction and that you only get particular quotients of the fundamental group; the ranks of these groups are lower bounds for the rank of the fundamental group itself. I think $I(mathop{#}_n K)geq n+1$ still applies, though.
$endgroup$
– Kyle Miller
Feb 7 at 0:05
add a comment |
$begingroup$
A handle decomposition of a ribbon disk gives a band presentation of the knot. If I understand how this goes correctly, the band presentation is an unlink along with attached bands between them such that
performing embedded arc surgery along these bands (which carry framing information for the surgery) gives the knot, and
if the unknots are thought of as vertices and the bands as edges, the graph forms a tree (since loops in the graph would correspond to a maxima, which don't exist).
The number of components in the unlink is the number of minima, and so your $I(K)$ is the minimal number of such components over all band presentations of the ribbon knot $K$.
For example, here is a band presentation of a square knot:
By the way, there appear to be some moves that can go between any two band presentations for the same ribbon disk of a ribbon knot:
(See at least Figure 3 of https://arxiv.org/abs/1804.09169 or the paper they cite:
Swenton, Frank J., On a calculus for 2-knots and surfaces in 4-space, J. Knot Theory Ramifications 10, No. 8, 1133-1141 (2001). ZBL1001.57044.). I wonder what additional moves you might need to go between any two ribbon disks.
Anyway, there is a nice way to calculate a group that is the homomorphic image of the fundamental group of the knot complement from such a band presentation, as you are probably aware:
Johnson, Dennis, Homomorphs of knot groups, Proc. Am. Math. Soc. 78, 135-138 (1980). ZBL0435.57003.
Each unknot is assigned a generator, and each band sum introduces a single additional relation depending on which unknots the band passes through. This implies that $I(K)geq operatorname{rank}(pi_1(S^3-K))$, where the rank is the minimal number of generators over all presentations of the finitely generated group $pi_1(S^3-K)$.
The following paper implies that there are knot groups of arbitrary rank:
Weidmann, Richard, On the rank of amalgamated products and product knot groups, Math. Ann. 312, No. 4, 761-771 (1998). ZBL0926.20019.
In particular, the $n$-fold connect sum of any nontrivial ribbon knot $K$ has $I(mathop{#}_nK)geq operatorname{rank}(pi_1(S^3-mathop{#}_nK))geq n+1$.
(If there were a way to modify Johnson's construction to generate prime knots, then one could get an infinite family of prime ribbon knots with arbitrary rank.)
answered Feb 4 at 1:48
Kyle MillerKyle Miller
9,700930
$endgroup$
A handle decomposition of a ribbon disk gives a band presentation of the knot. If I understand how this goes correctly, the band presentation is an unlink along with attached bands between them such that
performing embedded arc surgery along these bands (which carry framing information for the surgery) gives the knot, and
if the unknots are thought of as vertices and the bands as edges, the graph forms a tree (since loops in the graph would correspond to a maxima, which don't exist).
The number of components in the unlink is the number of minima, and so your $I(K)$ is the minimal number of such components over all band presentations of the ribbon knot $K$.
For example, here is a band presentation of a square knot:
By the way, there appear to be some moves that can go between any two band presentations for the same ribbon disk of a ribbon knot:
(See at least Figure 3 of https://arxiv.org/abs/1804.09169 or the paper they cite:
Swenton, Frank J., On a calculus for 2-knots and surfaces in 4-space, J. Knot Theory Ramifications 10, No. 8, 1133-1141 (2001). ZBL1001.57044.). I wonder what additional moves you might need to go between any two ribbon disks.
Anyway, there is a nice way to calculate a group that is the homomorphic image of the fundamental group of the knot complement from such a band presentation, as you are probably aware:
Johnson, Dennis, Homomorphs of knot groups, Proc. Am. Math. Soc. 78, 135-138 (1980). ZBL0435.57003.
Each unknot is assigned a generator, and each band sum introduces a single additional relation depending on which unknots the band passes through. This implies that $I(K)geq operatorname{rank}(pi_1(S^3-K))$, where the rank is the minimal number of generators over all presentations of the finitely generated group $pi_1(S^3-K)$.
The following paper implies that there are knot groups of arbitrary rank:
Weidmann, Richard, On the rank of amalgamated products and product knot groups, Math. Ann. 312, No. 4, 761-771 (1998). ZBL0926.20019.
In particular, the $n$-fold connect sum of any nontrivial ribbon knot $K$ has $I(mathop{#}_nK)geq operatorname{rank}(pi_1(S^3-mathop{#}_nK))geq n+1$.
(If there were a way to modify Johnson's construction to generate prime knots, then one could get an infinite family of prime ribbon knots with arbitrary rank.)
answered Feb 4 at 1:48
Kyle MillerKyle Miller
9,700930
edited Feb 5 at 18:47
answered Feb 4 at 1:48
Kyle MillerKyle Miller
9,700930
answered Feb 4 at 1:48
Kyle MillerKyle Miller
9,700930
answered Feb 4 at 1:48
Kyle MillerKyle Miller
9,700930
9,700930
1
$begingroup$
Thanks Kyle, this is great! I had forgotten about the Johnson paper and I did not know about that result of Weidmann - that is exactly the sort of fact that I was looking for!!
$endgroup$
– user101010
Feb 5 at 0:29
$begingroup$
@user101010 I've realized I misunderstood the kind of group presentation you get from Johnson's construction and that you only get particular quotients of the fundamental group; the ranks of these groups are lower bounds for the rank of the fundamental group itself. I think $I(mathop{#}_n K)geq n+1$ still applies, though.
$endgroup$
– Kyle Miller
Feb 7 at 0:05
add a comment |
1
$begingroup$
Thanks Kyle, this is great! I had forgotten about the Johnson paper and I did not know about that result of Weidmann - that is exactly the sort of fact that I was looking for!!
$endgroup$
– user101010
Feb 5 at 0:29
$begingroup$
@user101010 I've realized I misunderstood the kind of group presentation you get from Johnson's construction and that you only get particular quotients of the fundamental group; the ranks of these groups are lower bounds for the rank of the fundamental group itself. I think $I(mathop{#}_n K)geq n+1$ still applies, though.
$endgroup$
– Kyle Miller
Feb 7 at 0:05
1
1
$begingroup$
Thanks Kyle, this is great! I had forgotten about the Johnson paper and I did not know about that result of Weidmann - that is exactly the sort of fact that I was looking for!!
$endgroup$
– user101010
Feb 5 at 0:29
$begingroup$
Thanks Kyle, this is great! I had forgotten about the Johnson paper and I did not know about that result of Weidmann - that is exactly the sort of fact that I was looking for!!
$endgroup$
– user101010
Feb 5 at 0:29
$begingroup$
@user101010 I've realized I misunderstood the kind of group presentation you get from Johnson's construction and that you only get particular quotients of the fundamental group; the ranks of these groups are lower bounds for the rank of the fundamental group itself. I think $I(mathop{#}_n K)geq n+1$ still applies, though.
$endgroup$
– Kyle Miller
Feb 7 at 0:05
$begingroup$
@user101010 I've realized I misunderstood the kind of group presentation you get from Johnson's construction and that you only get particular quotients of the fundamental group; the ranks of these groups are lower bounds for the rank of the fundamental group itself. I think $I(mathop{#}_n K)geq n+1$ still applies, though.
$endgroup$
– Kyle Miller
Feb 7 at 0:05
add a comment |
$begingroup$
Since the Johnson paper only gives a quotient of the knot group, it is not clear why $I(#_n K)geq n$, and in particular, it is not clear if $I$ can be arbitrarily large. Am I missing something obvious?
answered Mar 22 at 21:16
Sucharit SarkarSucharit Sarkar
101
New contributor
Sucharit Sarkar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Since the Johnson paper only gives a quotient of the knot group, it is not clear why $I(#_n K)geq n$, and in particular, it is not clear if $I$ can be arbitrarily large. Am I missing something obvious?
answered Mar 22 at 21:16
Sucharit SarkarSucharit Sarkar
101
New contributor
Sucharit Sarkar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Since the Johnson paper only gives a quotient of the knot group, it is not clear why $I(#_n K)geq n$, and in particular, it is not clear if $I$ can be arbitrarily large. Am I missing something obvious?
answered Mar 22 at 21:16
Sucharit SarkarSucharit Sarkar
101
New contributor
Sucharit Sarkar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Since the Johnson paper only gives a quotient of the knot group, it is not clear why $I(#_n K)geq n$, and in particular, it is not clear if $I$ can be arbitrarily large. Am I missing something obvious?
answered Mar 22 at 21:16
Sucharit SarkarSucharit Sarkar
101
New contributor
Sucharit Sarkar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 22 at 21:16
Sucharit SarkarSucharit Sarkar
101
New contributor
Sucharit Sarkar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 22 at 21:16
Sucharit SarkarSucharit Sarkar
101
answered Mar 22 at 21:16
Sucharit SarkarSucharit Sarkar
101
101
New contributor
Sucharit Sarkar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sucharit Sarkar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sucharit Sarkar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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