Do i imagine the linear (straight line) homotopy in a correct way? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)simple proposition about homotopy groupsymmetry in the homotopy relationDetermining images of points in a path homotopy.Difference between two concepts of homotopy for simplicial maps?Maps to Sn homotopicShow that g*c and c*g are homotopic, where g is a loop and c a constant loopComputing fundamental group of the complement to three infinite straight lines, and of complement to $S^1 cup Z $Linking numbers.tensor product and loop space multiplication are homotopic on $mathbbP^infty$Star convex subset $A$ of $mathbbR^n$ is simply connected
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Do i imagine the linear (straight line) homotopy in a correct way?
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Do i imagine the linear (straight line) homotopy in a correct way?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)simple proposition about homotopy groupsymmetry in the homotopy relationDetermining images of points in a path homotopy.Difference between two concepts of homotopy for simplicial maps?Maps to Sn homotopicShow that g*c and c*g are homotopic, where g is a loop and c a constant loopComputing fundamental group of the complement to three infinite straight lines, and of complement to $S^1 cup Z $Linking numbers.tensor product and loop space multiplication are homotopic on $mathbbP^infty$Star convex subset $A$ of $mathbbR^n$ is simply connected
$begingroup$
Today i learned about the linear homotopy which says that any two paths $f_0, f_1$ in $mathbbR^n$ are homotopic via the homotopy $$ f_t(s) = (1-t)f_0(s) + tf_1(s)$$
Am i right in imagining the given homotopy as something like this?
such that $F(s,t) = f_t(s) $ are simply the linesegments going from $f_0(s)$ towards $f_1(s)$ as the straight lines (which "connect" $f_0$ and $f_1$ for every $sin [0,1]$) as drawn in the picture?
Sorry if this question might be a trivial one, i just want to make sure i don't get things wrong.
Thanks for any kind of feedback!
algebraic-topology homotopy-theory path-connected
asked 1 hour ago
ZestZest
301213
$endgroup$
add a comment |
$begingroup$
Today i learned about the linear homotopy which says that any two paths $f_0, f_1$ in $mathbbR^n$ are homotopic via the homotopy $$ f_t(s) = (1-t)f_0(s) + tf_1(s)$$
Am i right in imagining the given homotopy as something like this?
such that $F(s,t) = f_t(s) $ are simply the linesegments going from $f_0(s)$ towards $f_1(s)$ as the straight lines (which "connect" $f_0$ and $f_1$ for every $sin [0,1]$) as drawn in the picture?
Sorry if this question might be a trivial one, i just want to make sure i don't get things wrong.
Thanks for any kind of feedback!
algebraic-topology homotopy-theory path-connected
asked 1 hour ago
ZestZest
301213
$endgroup$
1
$begingroup$
It's correct. Furthermore, any point $s$ moves along its straight line at constant speed.
$endgroup$
– Lukas Kofler
1 hour ago
$begingroup$
isn't the point $t$ rather moving along its straight line? to me, $s$ is the parameter moving along the lines from $x_0$ to $x_1$ whereas $t$ moves along the straight linesegments between $f_0$ and $f_1$
$endgroup$
– Zest
1 hour ago
1
$begingroup$
That's what I was trying to say, sorry -- a fixed point $f_0(s)$ moves at constant speed as $t$ varies.
$endgroup$
– Lukas Kofler
1 hour ago
add a comment |
$begingroup$
Today i learned about the linear homotopy which says that any two paths $f_0, f_1$ in $mathbbR^n$ are homotopic via the homotopy $$ f_t(s) = (1-t)f_0(s) + tf_1(s)$$
Am i right in imagining the given homotopy as something like this?
such that $F(s,t) = f_t(s) $ are simply the linesegments going from $f_0(s)$ towards $f_1(s)$ as the straight lines (which "connect" $f_0$ and $f_1$ for every $sin [0,1]$) as drawn in the picture?
Sorry if this question might be a trivial one, i just want to make sure i don't get things wrong.
Thanks for any kind of feedback!
algebraic-topology homotopy-theory path-connected
asked 1 hour ago
ZestZest
301213
$endgroup$
Today i learned about the linear homotopy which says that any two paths $f_0, f_1$ in $mathbbR^n$ are homotopic via the homotopy $$ f_t(s) = (1-t)f_0(s) + tf_1(s)$$
Am i right in imagining the given homotopy as something like this?
such that $F(s,t) = f_t(s) $ are simply the linesegments going from $f_0(s)$ towards $f_1(s)$ as the straight lines (which "connect" $f_0$ and $f_1$ for every $sin [0,1]$) as drawn in the picture?
Sorry if this question might be a trivial one, i just want to make sure i don't get things wrong.
Thanks for any kind of feedback!
algebraic-topology homotopy-theory path-connected
algebraic-topology homotopy-theory path-connected
asked 1 hour ago
ZestZest
301213
asked 1 hour ago
ZestZest
301213
asked 1 hour ago
ZestZest
301213
asked 1 hour ago
ZestZest
301213
asked 1 hour ago
ZestZest
301213
301213
1
$begingroup$
It's correct. Furthermore, any point $s$ moves along its straight line at constant speed.
$endgroup$
– Lukas Kofler
1 hour ago
$begingroup$
isn't the point $t$ rather moving along its straight line? to me, $s$ is the parameter moving along the lines from $x_0$ to $x_1$ whereas $t$ moves along the straight linesegments between $f_0$ and $f_1$
$endgroup$
– Zest
1 hour ago
1
$begingroup$
That's what I was trying to say, sorry -- a fixed point $f_0(s)$ moves at constant speed as $t$ varies.
$endgroup$
– Lukas Kofler
1 hour ago
add a comment |
1
$begingroup$
It's correct. Furthermore, any point $s$ moves along its straight line at constant speed.
$endgroup$
– Lukas Kofler
1 hour ago
$begingroup$
isn't the point $t$ rather moving along its straight line? to me, $s$ is the parameter moving along the lines from $x_0$ to $x_1$ whereas $t$ moves along the straight linesegments between $f_0$ and $f_1$
$endgroup$
– Zest
1 hour ago
1
$begingroup$
That's what I was trying to say, sorry -- a fixed point $f_0(s)$ moves at constant speed as $t$ varies.
$endgroup$
– Lukas Kofler
1 hour ago
1
1
$begingroup$
It's correct. Furthermore, any point $s$ moves along its straight line at constant speed.
$endgroup$
– Lukas Kofler
1 hour ago
$begingroup$
It's correct. Furthermore, any point $s$ moves along its straight line at constant speed.
$endgroup$
– Lukas Kofler
1 hour ago
$begingroup$
isn't the point $t$ rather moving along its straight line? to me, $s$ is the parameter moving along the lines from $x_0$ to $x_1$ whereas $t$ moves along the straight linesegments between $f_0$ and $f_1$
$endgroup$
– Zest
1 hour ago
$begingroup$
isn't the point $t$ rather moving along its straight line? to me, $s$ is the parameter moving along the lines from $x_0$ to $x_1$ whereas $t$ moves along the straight linesegments between $f_0$ and $f_1$
$endgroup$
– Zest
1 hour ago
1
1
$begingroup$
That's what I was trying to say, sorry -- a fixed point $f_0(s)$ moves at constant speed as $t$ varies.
$endgroup$
– Lukas Kofler
1 hour ago
$begingroup$
That's what I was trying to say, sorry -- a fixed point $f_0(s)$ moves at constant speed as $t$ varies.
$endgroup$
– Lukas Kofler
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes. Perhaps to help you see why this is true, pick an arbitrary $s$ value $bars$ and call $a := f_0(bars)$ and $b := f_1(bars)$. Examining the homotopy,
$$
H_bars(t) := F(bars, t) = a(1-t) + bt
$$
we see that it is the parametric equation of a straight line connecting points $a$ and $b$ in $mathbbR^n$. If you draw this line for each $bars$ choice, you get your diagram. Perhaps write a bit of code to construct such a plot?
Furthermore, one may be interested in the "speed" at which $a$ "moves" to $b$. We find it to be a constant,
$$
dfracdH_barsdt = b - a
$$
answered 1 hour ago
jnez71jnez71
2,495720
$endgroup$
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
add a comment |
$begingroup$
Yes, it is absolutely correct.
$endgroup$
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
|
show 1 more comment
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes. Perhaps to help you see why this is true, pick an arbitrary $s$ value $bars$ and call $a := f_0(bars)$ and $b := f_1(bars)$. Examining the homotopy,
$$
H_bars(t) := F(bars, t) = a(1-t) + bt
$$
we see that it is the parametric equation of a straight line connecting points $a$ and $b$ in $mathbbR^n$. If you draw this line for each $bars$ choice, you get your diagram. Perhaps write a bit of code to construct such a plot?
Furthermore, one may be interested in the "speed" at which $a$ "moves" to $b$. We find it to be a constant,
$$
dfracdH_barsdt = b - a
$$
answered 1 hour ago
jnez71jnez71
2,495720
$endgroup$
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
add a comment |
$begingroup$
Yes. Perhaps to help you see why this is true, pick an arbitrary $s$ value $bars$ and call $a := f_0(bars)$ and $b := f_1(bars)$. Examining the homotopy,
$$
H_bars(t) := F(bars, t) = a(1-t) + bt
$$
we see that it is the parametric equation of a straight line connecting points $a$ and $b$ in $mathbbR^n$. If you draw this line for each $bars$ choice, you get your diagram. Perhaps write a bit of code to construct such a plot?
Furthermore, one may be interested in the "speed" at which $a$ "moves" to $b$. We find it to be a constant,
$$
dfracdH_barsdt = b - a
$$
answered 1 hour ago
jnez71jnez71
2,495720
$endgroup$
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
add a comment |
$begingroup$
Yes. Perhaps to help you see why this is true, pick an arbitrary $s$ value $bars$ and call $a := f_0(bars)$ and $b := f_1(bars)$. Examining the homotopy,
$$
H_bars(t) := F(bars, t) = a(1-t) + bt
$$
we see that it is the parametric equation of a straight line connecting points $a$ and $b$ in $mathbbR^n$. If you draw this line for each $bars$ choice, you get your diagram. Perhaps write a bit of code to construct such a plot?
Furthermore, one may be interested in the "speed" at which $a$ "moves" to $b$. We find it to be a constant,
$$
dfracdH_barsdt = b - a
$$
answered 1 hour ago
jnez71jnez71
2,495720
$endgroup$
Yes. Perhaps to help you see why this is true, pick an arbitrary $s$ value $bars$ and call $a := f_0(bars)$ and $b := f_1(bars)$. Examining the homotopy,
$$
H_bars(t) := F(bars, t) = a(1-t) + bt
$$
we see that it is the parametric equation of a straight line connecting points $a$ and $b$ in $mathbbR^n$. If you draw this line for each $bars$ choice, you get your diagram. Perhaps write a bit of code to construct such a plot?
Furthermore, one may be interested in the "speed" at which $a$ "moves" to $b$. We find it to be a constant,
$$
dfracdH_barsdt = b - a
$$
answered 1 hour ago
jnez71jnez71
2,495720
edited 1 hour ago
answered 1 hour ago
jnez71jnez71
2,495720
answered 1 hour ago
jnez71jnez71
2,495720
answered 1 hour ago
jnez71jnez71
2,495720
2,495720
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
add a comment |
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
1
1
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
add a comment |
$begingroup$
Yes, it is absolutely correct.
$endgroup$
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
|
show 1 more comment
$begingroup$
Yes, it is absolutely correct.
$endgroup$
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
|
show 1 more comment
$begingroup$
Yes, it is absolutely correct.
$endgroup$
Yes, it is absolutely correct.
answered 1 hour ago
community wiki
Paul Frost
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
|
show 1 more comment
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
1
1
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
|
show 1 more comment
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It's correct. Furthermore, any point $s$ moves along its straight line at constant speed.
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– Lukas Kofler
1 hour ago
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isn't the point $t$ rather moving along its straight line? to me, $s$ is the parameter moving along the lines from $x_0$ to $x_1$ whereas $t$ moves along the straight linesegments between $f_0$ and $f_1$
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– Zest
1 hour ago
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$begingroup$
That's what I was trying to say, sorry -- a fixed point $f_0(s)$ moves at constant speed as $t$ varies.
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– Lukas Kofler
1 hour ago