Explanations about the volume of a regular simplex












2














I'm really sorry, this may sound ridiculous but I can't understand the Wikipedia explanation about the volume of regular n-dimensional simplices, here.



In particular, these two sentences make no sense to me:




If the coordinates of a point in a unit n-box are sorted, together
with 0 and 1, and successive differences are taken, then since the
results add to one, the result is a point in an n simplex spanned by
the origin and the closest n vertices of the box. The taking of
differences was a unimodular (volume-preserving) transformation, but
sorting compressed the space by a factor of n!.




I think this might relate to the section about increasing coordinates (although I can't exactly see how), which I mostly understood but then again this sentence about volume measurement is obscure to me as well:




Indeed, the ordered simplex is a (closed) fundamental domain for the
action of the symmetric group on the n-cube, meaning that the orbit of
the ordered simplex under the n! elements of the symmetric group
divides the n-cube into n! mostly disjoint simplices (disjoint except
for boundaries), showing that this simplex has volume 1/n!.




I'm trying to read about fundamental domains to shed some light on the previous sentence, but it's not the easiest subject.. Help would be muchly appreciated.






volume simplex







share|cite|improve this question






















  • Yes, that is a confusing first sentence, at least taken out of context. It begins by referring to one point in the "unit box" and somehow asserts $n$ or more "vertices of the box", used (together with the origin) to define (span) an $n$-simplex.
    – hardmath
    Nov 18 '14 at 11:43










  • I guess the point I find non-trivial is since the results add to one. If this is true, then the point in question is indeed on the standard n-simplex in $mathbb{R}^{n+1}$ (which is a region of the hyperplane $sum_i x_i = 1$). But then I don't understand closest n vertices on the box either, since these vertices would always the same; they correspond to the canonical vectors. The n-box has $2^n$ vertices, why introduce such a great uncertainty in the choice of "closest vertices"? Finally, where does but sorting compressed the space by a factor of n! come from?
    – Sheljohn
    Nov 18 '14 at 11:55








  • 1




    Taking a point $(x_1,x_2,ldots,x_n)$ "in" the unit box means $0 le x_1,x_2,ldots,x_n le 1$. If the coordinates were sorted (ascending), so that $0 le x_1 le x_2 ;ldots le x_n le 1$, then taking differences $y_i = x_{i+1} - x_i$, where for convenience we take $x_0 = 0$, $x_{n+1} = 1$, we get $y_i in [0,1], i = 0,ldots,n$ such that $sum y_i = 1$.
    – hardmath
    Nov 18 '14 at 12:45










  • @hardmath You're right, it's just a telescopic sum. Thank you! Well, first point checked then :)
    – Sheljohn
    Nov 18 '14 at 13:31










  • It seems the idea is to present a map of the "unit box" in $mathbb{R}^n$ onto the standard simplex, and show that it has a nice interpretation as a $n!$ to $1$ mapping that preserves volume on portions where the coordinates are already sorted. I'll have to puzzle over how to present the idea more clearly.
    – hardmath
    Nov 18 '14 at 13:37


















2














I'm really sorry, this may sound ridiculous but I can't understand the Wikipedia explanation about the volume of regular n-dimensional simplices, here.



In particular, these two sentences make no sense to me:




If the coordinates of a point in a unit n-box are sorted, together
with 0 and 1, and successive differences are taken, then since the
results add to one, the result is a point in an n simplex spanned by
the origin and the closest n vertices of the box. The taking of
differences was a unimodular (volume-preserving) transformation, but
sorting compressed the space by a factor of n!.




I think this might relate to the section about increasing coordinates (although I can't exactly see how), which I mostly understood but then again this sentence about volume measurement is obscure to me as well:




Indeed, the ordered simplex is a (closed) fundamental domain for the
action of the symmetric group on the n-cube, meaning that the orbit of
the ordered simplex under the n! elements of the symmetric group
divides the n-cube into n! mostly disjoint simplices (disjoint except
for boundaries), showing that this simplex has volume 1/n!.




I'm trying to read about fundamental domains to shed some light on the previous sentence, but it's not the easiest subject.. Help would be muchly appreciated.






volume simplex







share|cite|improve this question






















  • Yes, that is a confusing first sentence, at least taken out of context. It begins by referring to one point in the "unit box" and somehow asserts $n$ or more "vertices of the box", used (together with the origin) to define (span) an $n$-simplex.
    – hardmath
    Nov 18 '14 at 11:43










  • I guess the point I find non-trivial is since the results add to one. If this is true, then the point in question is indeed on the standard n-simplex in $mathbb{R}^{n+1}$ (which is a region of the hyperplane $sum_i x_i = 1$). But then I don't understand closest n vertices on the box either, since these vertices would always the same; they correspond to the canonical vectors. The n-box has $2^n$ vertices, why introduce such a great uncertainty in the choice of "closest vertices"? Finally, where does but sorting compressed the space by a factor of n! come from?
    – Sheljohn
    Nov 18 '14 at 11:55








  • 1




    Taking a point $(x_1,x_2,ldots,x_n)$ "in" the unit box means $0 le x_1,x_2,ldots,x_n le 1$. If the coordinates were sorted (ascending), so that $0 le x_1 le x_2 ;ldots le x_n le 1$, then taking differences $y_i = x_{i+1} - x_i$, where for convenience we take $x_0 = 0$, $x_{n+1} = 1$, we get $y_i in [0,1], i = 0,ldots,n$ such that $sum y_i = 1$.
    – hardmath
    Nov 18 '14 at 12:45










  • @hardmath You're right, it's just a telescopic sum. Thank you! Well, first point checked then :)
    – Sheljohn
    Nov 18 '14 at 13:31










  • It seems the idea is to present a map of the "unit box" in $mathbb{R}^n$ onto the standard simplex, and show that it has a nice interpretation as a $n!$ to $1$ mapping that preserves volume on portions where the coordinates are already sorted. I'll have to puzzle over how to present the idea more clearly.
    – hardmath
    Nov 18 '14 at 13:37
















2












2








2







I'm really sorry, this may sound ridiculous but I can't understand the Wikipedia explanation about the volume of regular n-dimensional simplices, here.



In particular, these two sentences make no sense to me:




If the coordinates of a point in a unit n-box are sorted, together
with 0 and 1, and successive differences are taken, then since the
results add to one, the result is a point in an n simplex spanned by
the origin and the closest n vertices of the box. The taking of
differences was a unimodular (volume-preserving) transformation, but
sorting compressed the space by a factor of n!.




I think this might relate to the section about increasing coordinates (although I can't exactly see how), which I mostly understood but then again this sentence about volume measurement is obscure to me as well:




Indeed, the ordered simplex is a (closed) fundamental domain for the
action of the symmetric group on the n-cube, meaning that the orbit of
the ordered simplex under the n! elements of the symmetric group
divides the n-cube into n! mostly disjoint simplices (disjoint except
for boundaries), showing that this simplex has volume 1/n!.




I'm trying to read about fundamental domains to shed some light on the previous sentence, but it's not the easiest subject.. Help would be muchly appreciated.






volume simplex







share|cite|improve this question













I'm really sorry, this may sound ridiculous but I can't understand the Wikipedia explanation about the volume of regular n-dimensional simplices, here.



In particular, these two sentences make no sense to me:




If the coordinates of a point in a unit n-box are sorted, together
with 0 and 1, and successive differences are taken, then since the
results add to one, the result is a point in an n simplex spanned by
the origin and the closest n vertices of the box. The taking of
differences was a unimodular (volume-preserving) transformation, but
sorting compressed the space by a factor of n!.




I think this might relate to the section about increasing coordinates (although I can't exactly see how), which I mostly understood but then again this sentence about volume measurement is obscure to me as well:




Indeed, the ordered simplex is a (closed) fundamental domain for the
action of the symmetric group on the n-cube, meaning that the orbit of
the ordered simplex under the n! elements of the symmetric group
divides the n-cube into n! mostly disjoint simplices (disjoint except
for boundaries), showing that this simplex has volume 1/n!.




I'm trying to read about fundamental domains to shed some light on the previous sentence, but it's not the easiest subject.. Help would be muchly appreciated.






volume simplex




volume simplex






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 18 '14 at 11:28









Sheljohn

1,267722




1,267722












  • Yes, that is a confusing first sentence, at least taken out of context. It begins by referring to one point in the "unit box" and somehow asserts $n$ or more "vertices of the box", used (together with the origin) to define (span) an $n$-simplex.
    – hardmath
    Nov 18 '14 at 11:43










  • I guess the point I find non-trivial is since the results add to one. If this is true, then the point in question is indeed on the standard n-simplex in $mathbb{R}^{n+1}$ (which is a region of the hyperplane $sum_i x_i = 1$). But then I don't understand closest n vertices on the box either, since these vertices would always the same; they correspond to the canonical vectors. The n-box has $2^n$ vertices, why introduce such a great uncertainty in the choice of "closest vertices"? Finally, where does but sorting compressed the space by a factor of n! come from?
    – Sheljohn
    Nov 18 '14 at 11:55








  • 1




    Taking a point $(x_1,x_2,ldots,x_n)$ "in" the unit box means $0 le x_1,x_2,ldots,x_n le 1$. If the coordinates were sorted (ascending), so that $0 le x_1 le x_2 ;ldots le x_n le 1$, then taking differences $y_i = x_{i+1} - x_i$, where for convenience we take $x_0 = 0$, $x_{n+1} = 1$, we get $y_i in [0,1], i = 0,ldots,n$ such that $sum y_i = 1$.
    – hardmath
    Nov 18 '14 at 12:45










  • @hardmath You're right, it's just a telescopic sum. Thank you! Well, first point checked then :)
    – Sheljohn
    Nov 18 '14 at 13:31










  • It seems the idea is to present a map of the "unit box" in $mathbb{R}^n$ onto the standard simplex, and show that it has a nice interpretation as a $n!$ to $1$ mapping that preserves volume on portions where the coordinates are already sorted. I'll have to puzzle over how to present the idea more clearly.
    – hardmath
    Nov 18 '14 at 13:37




















  • Yes, that is a confusing first sentence, at least taken out of context. It begins by referring to one point in the "unit box" and somehow asserts $n$ or more "vertices of the box", used (together with the origin) to define (span) an $n$-simplex.
    – hardmath
    Nov 18 '14 at 11:43










  • I guess the point I find non-trivial is since the results add to one. If this is true, then the point in question is indeed on the standard n-simplex in $mathbb{R}^{n+1}$ (which is a region of the hyperplane $sum_i x_i = 1$). But then I don't understand closest n vertices on the box either, since these vertices would always the same; they correspond to the canonical vectors. The n-box has $2^n$ vertices, why introduce such a great uncertainty in the choice of "closest vertices"? Finally, where does but sorting compressed the space by a factor of n! come from?
    – Sheljohn
    Nov 18 '14 at 11:55








  • 1




    Taking a point $(x_1,x_2,ldots,x_n)$ "in" the unit box means $0 le x_1,x_2,ldots,x_n le 1$. If the coordinates were sorted (ascending), so that $0 le x_1 le x_2 ;ldots le x_n le 1$, then taking differences $y_i = x_{i+1} - x_i$, where for convenience we take $x_0 = 0$, $x_{n+1} = 1$, we get $y_i in [0,1], i = 0,ldots,n$ such that $sum y_i = 1$.
    – hardmath
    Nov 18 '14 at 12:45










  • @hardmath You're right, it's just a telescopic sum. Thank you! Well, first point checked then :)
    – Sheljohn
    Nov 18 '14 at 13:31










  • It seems the idea is to present a map of the "unit box" in $mathbb{R}^n$ onto the standard simplex, and show that it has a nice interpretation as a $n!$ to $1$ mapping that preserves volume on portions where the coordinates are already sorted. I'll have to puzzle over how to present the idea more clearly.
    – hardmath
    Nov 18 '14 at 13:37


















Yes, that is a confusing first sentence, at least taken out of context. It begins by referring to one point in the "unit box" and somehow asserts $n$ or more "vertices of the box", used (together with the origin) to define (span) an $n$-simplex.
– hardmath
Nov 18 '14 at 11:43




Yes, that is a confusing first sentence, at least taken out of context. It begins by referring to one point in the "unit box" and somehow asserts $n$ or more "vertices of the box", used (together with the origin) to define (span) an $n$-simplex.
– hardmath
Nov 18 '14 at 11:43












I guess the point I find non-trivial is since the results add to one. If this is true, then the point in question is indeed on the standard n-simplex in $mathbb{R}^{n+1}$ (which is a region of the hyperplane $sum_i x_i = 1$). But then I don't understand closest n vertices on the box either, since these vertices would always the same; they correspond to the canonical vectors. The n-box has $2^n$ vertices, why introduce such a great uncertainty in the choice of "closest vertices"? Finally, where does but sorting compressed the space by a factor of n! come from?
– Sheljohn
Nov 18 '14 at 11:55






I guess the point I find non-trivial is since the results add to one. If this is true, then the point in question is indeed on the standard n-simplex in $mathbb{R}^{n+1}$ (which is a region of the hyperplane $sum_i x_i = 1$). But then I don't understand closest n vertices on the box either, since these vertices would always the same; they correspond to the canonical vectors. The n-box has $2^n$ vertices, why introduce such a great uncertainty in the choice of "closest vertices"? Finally, where does but sorting compressed the space by a factor of n! come from?
– Sheljohn
Nov 18 '14 at 11:55






1




1




Taking a point $(x_1,x_2,ldots,x_n)$ "in" the unit box means $0 le x_1,x_2,ldots,x_n le 1$. If the coordinates were sorted (ascending), so that $0 le x_1 le x_2 ;ldots le x_n le 1$, then taking differences $y_i = x_{i+1} - x_i$, where for convenience we take $x_0 = 0$, $x_{n+1} = 1$, we get $y_i in [0,1], i = 0,ldots,n$ such that $sum y_i = 1$.
– hardmath
Nov 18 '14 at 12:45




Taking a point $(x_1,x_2,ldots,x_n)$ "in" the unit box means $0 le x_1,x_2,ldots,x_n le 1$. If the coordinates were sorted (ascending), so that $0 le x_1 le x_2 ;ldots le x_n le 1$, then taking differences $y_i = x_{i+1} - x_i$, where for convenience we take $x_0 = 0$, $x_{n+1} = 1$, we get $y_i in [0,1], i = 0,ldots,n$ such that $sum y_i = 1$.
– hardmath
Nov 18 '14 at 12:45












@hardmath You're right, it's just a telescopic sum. Thank you! Well, first point checked then :)
– Sheljohn
Nov 18 '14 at 13:31




@hardmath You're right, it's just a telescopic sum. Thank you! Well, first point checked then :)
– Sheljohn
Nov 18 '14 at 13:31












It seems the idea is to present a map of the "unit box" in $mathbb{R}^n$ onto the standard simplex, and show that it has a nice interpretation as a $n!$ to $1$ mapping that preserves volume on portions where the coordinates are already sorted. I'll have to puzzle over how to present the idea more clearly.
– hardmath
Nov 18 '14 at 13:37






It seems the idea is to present a map of the "unit box" in $mathbb{R}^n$ onto the standard simplex, and show that it has a nice interpretation as a $n!$ to $1$ mapping that preserves volume on portions where the coordinates are already sorted. I'll have to puzzle over how to present the idea more clearly.
– hardmath
Nov 18 '14 at 13:37












1 Answer
1






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oldest

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You might check the proof of



Cartesian coordinates for vertices of a regular 16-simplex?



for a formula for the regular simplex volume.






share|cite|improve this answer























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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    You might check the proof of



    Cartesian coordinates for vertices of a regular 16-simplex?



    for a formula for the regular simplex volume.






    share|cite|improve this answer




























      0














      You might check the proof of



      Cartesian coordinates for vertices of a regular 16-simplex?



      for a formula for the regular simplex volume.






      share|cite|improve this answer


























        0












        0








        0






        You might check the proof of



        Cartesian coordinates for vertices of a regular 16-simplex?



        for a formula for the regular simplex volume.






        share|cite|improve this answer














        You might check the proof of



        Cartesian coordinates for vertices of a regular 16-simplex?



        for a formula for the regular simplex volume.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 13 '17 at 12:21









        Community♦

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        1










        answered Jul 27 '15 at 21:11









        andre

        1,262412




        1,262412






























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