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Trying to prove $partial^2=0$ on $k$-cells
Clash Royale CLAN TAG#URR8PPP
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Consider a $k$-cell $ c : [0,1]^k to U subset mathbbR^n , (t_1,...,t_k) mapsto c (t_1,...,t_k) $. Then the boundary of $c$, $partial c$ is defined as
$$ partial c := sum_i=1^k (-1)^i-1 left( c_i^1 - c_i^0 right) $$
where
$$ c_i^1 := c(t_1,...,t_i-1,1,t_i+1,...,t_k) \ c_i^0 := c(t_1,...,t_i-1,0,t_i+1,...,t_k) $$
Then supposing $c = sum_m a^m c_m$ is a $k$-chain ($a^m in mathbbR$, $c_m$ $k$-cells), it’s boundary is defined by
$$ partial c = partial left( sum_m a^m c_m right) := sum_m a^m partial c_m $$
I’m struggling to prove the special property of $partial$ that $partial^2 = 0$. Here’s what I have so far...
Since $partial$ is linear, it is sufficient to prove $partial^2 c = 0$ for any $k$-cell $c$.
Introducing another sigma sign, the definition of $partial c$ can be made to look nicer,
$$
partial c := sum_i=1^k (-1)^i-1 left( c_i^1 - c_i^0 right)
\
= sum_i=1^k (-1)^i-1 sum_rho=0^1 (-1)^rho+1 c_i^rho \
= sum_i=1^k sum_rho=0^1 (-1)^i+rho c_i^rho
$$
Then, I tried brute forcing some algebra and definitions...
$$
partial^2 c = partial sum_i=1^k sum_rho=0^1 (-1)^i+rho c_i^rho
= sum_i=1^k sum_rho=0^1 (-1)^i+rho partial c_i^rho
\
= sum_i=1^k sum_rho=0^1 (-1)^i+rho sum_j=1 ^k sum_sigma=0^1 (-1)^j+sigma c_ij^rho sigma
\
= sum_ ineq j ^k sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c_ij^rhosigma
$$
where
$$
c_ij^rhosigma = c(t_1, ... , t_i-1 , rho , t_i+1 , ... , t_j-1 , sigma, t_j+1, ... , t_k)
$$
This is where I’m stuck.
differential-geometry
asked Jul 17 at 8:38
user577413
838
add a comment |Â
up vote
3
down vote
favorite
Consider a $k$-cell $ c : [0,1]^k to U subset mathbbR^n , (t_1,...,t_k) mapsto c (t_1,...,t_k) $. Then the boundary of $c$, $partial c$ is defined as
$$ partial c := sum_i=1^k (-1)^i-1 left( c_i^1 - c_i^0 right) $$
where
$$ c_i^1 := c(t_1,...,t_i-1,1,t_i+1,...,t_k) \ c_i^0 := c(t_1,...,t_i-1,0,t_i+1,...,t_k) $$
Then supposing $c = sum_m a^m c_m$ is a $k$-chain ($a^m in mathbbR$, $c_m$ $k$-cells), it’s boundary is defined by
$$ partial c = partial left( sum_m a^m c_m right) := sum_m a^m partial c_m $$
I’m struggling to prove the special property of $partial$ that $partial^2 = 0$. Here’s what I have so far...
Since $partial$ is linear, it is sufficient to prove $partial^2 c = 0$ for any $k$-cell $c$.
Introducing another sigma sign, the definition of $partial c$ can be made to look nicer,
$$
partial c := sum_i=1^k (-1)^i-1 left( c_i^1 - c_i^0 right)
\
= sum_i=1^k (-1)^i-1 sum_rho=0^1 (-1)^rho+1 c_i^rho \
= sum_i=1^k sum_rho=0^1 (-1)^i+rho c_i^rho
$$
Then, I tried brute forcing some algebra and definitions...
$$
partial^2 c = partial sum_i=1^k sum_rho=0^1 (-1)^i+rho c_i^rho
= sum_i=1^k sum_rho=0^1 (-1)^i+rho partial c_i^rho
\
= sum_i=1^k sum_rho=0^1 (-1)^i+rho sum_j=1 ^k sum_sigma=0^1 (-1)^j+sigma c_ij^rho sigma
\
= sum_ ineq j ^k sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c_ij^rhosigma
$$
where
$$
c_ij^rhosigma = c(t_1, ... , t_i-1 , rho , t_i+1 , ... , t_j-1 , sigma, t_j+1, ... , t_k)
$$
This is where I’m stuck.
differential-geometry
asked Jul 17 at 8:38
user577413
838
Try to do the calculations in some lower dimensions (e.g. $n=1,2,3$) first and without using sum signs.
– md2perpe
Jul 17 at 11:23
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Consider a $k$-cell $ c : [0,1]^k to U subset mathbbR^n , (t_1,...,t_k) mapsto c (t_1,...,t_k) $. Then the boundary of $c$, $partial c$ is defined as
$$ partial c := sum_i=1^k (-1)^i-1 left( c_i^1 - c_i^0 right) $$
where
$$ c_i^1 := c(t_1,...,t_i-1,1,t_i+1,...,t_k) \ c_i^0 := c(t_1,...,t_i-1,0,t_i+1,...,t_k) $$
Then supposing $c = sum_m a^m c_m$ is a $k$-chain ($a^m in mathbbR$, $c_m$ $k$-cells), it’s boundary is defined by
$$ partial c = partial left( sum_m a^m c_m right) := sum_m a^m partial c_m $$
I’m struggling to prove the special property of $partial$ that $partial^2 = 0$. Here’s what I have so far...
Since $partial$ is linear, it is sufficient to prove $partial^2 c = 0$ for any $k$-cell $c$.
Introducing another sigma sign, the definition of $partial c$ can be made to look nicer,
$$
partial c := sum_i=1^k (-1)^i-1 left( c_i^1 - c_i^0 right)
\
= sum_i=1^k (-1)^i-1 sum_rho=0^1 (-1)^rho+1 c_i^rho \
= sum_i=1^k sum_rho=0^1 (-1)^i+rho c_i^rho
$$
Then, I tried brute forcing some algebra and definitions...
$$
partial^2 c = partial sum_i=1^k sum_rho=0^1 (-1)^i+rho c_i^rho
= sum_i=1^k sum_rho=0^1 (-1)^i+rho partial c_i^rho
\
= sum_i=1^k sum_rho=0^1 (-1)^i+rho sum_j=1 ^k sum_sigma=0^1 (-1)^j+sigma c_ij^rho sigma
\
= sum_ ineq j ^k sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c_ij^rhosigma
$$
where
$$
c_ij^rhosigma = c(t_1, ... , t_i-1 , rho , t_i+1 , ... , t_j-1 , sigma, t_j+1, ... , t_k)
$$
This is where I’m stuck.
differential-geometry
asked Jul 17 at 8:38
user577413
838
Consider a $k$-cell $ c : [0,1]^k to U subset mathbbR^n , (t_1,...,t_k) mapsto c (t_1,...,t_k) $. Then the boundary of $c$, $partial c$ is defined as
$$ partial c := sum_i=1^k (-1)^i-1 left( c_i^1 - c_i^0 right) $$
where
$$ c_i^1 := c(t_1,...,t_i-1,1,t_i+1,...,t_k) \ c_i^0 := c(t_1,...,t_i-1,0,t_i+1,...,t_k) $$
Then supposing $c = sum_m a^m c_m$ is a $k$-chain ($a^m in mathbbR$, $c_m$ $k$-cells), it’s boundary is defined by
$$ partial c = partial left( sum_m a^m c_m right) := sum_m a^m partial c_m $$
I’m struggling to prove the special property of $partial$ that $partial^2 = 0$. Here’s what I have so far...
Since $partial$ is linear, it is sufficient to prove $partial^2 c = 0$ for any $k$-cell $c$.
Introducing another sigma sign, the definition of $partial c$ can be made to look nicer,
$$
partial c := sum_i=1^k (-1)^i-1 left( c_i^1 - c_i^0 right)
\
= sum_i=1^k (-1)^i-1 sum_rho=0^1 (-1)^rho+1 c_i^rho \
= sum_i=1^k sum_rho=0^1 (-1)^i+rho c_i^rho
$$
Then, I tried brute forcing some algebra and definitions...
$$
partial^2 c = partial sum_i=1^k sum_rho=0^1 (-1)^i+rho c_i^rho
= sum_i=1^k sum_rho=0^1 (-1)^i+rho partial c_i^rho
\
= sum_i=1^k sum_rho=0^1 (-1)^i+rho sum_j=1 ^k sum_sigma=0^1 (-1)^j+sigma c_ij^rho sigma
\
= sum_ ineq j ^k sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c_ij^rhosigma
$$
where
$$
c_ij^rhosigma = c(t_1, ... , t_i-1 , rho , t_i+1 , ... , t_j-1 , sigma, t_j+1, ... , t_k)
$$
This is where I’m stuck.
differential-geometry
asked Jul 17 at 8:38
user577413
838
asked Jul 17 at 8:38
user577413
838
asked Jul 17 at 8:38
user577413
838
asked Jul 17 at 8:38
user577413
838
838
Try to do the calculations in some lower dimensions (e.g. $n=1,2,3$) first and without using sum signs.
– md2perpe
Jul 17 at 11:23
add a comment |Â
Try to do the calculations in some lower dimensions (e.g. $n=1,2,3$) first and without using sum signs.
– md2perpe
Jul 17 at 11:23
Try to do the calculations in some lower dimensions (e.g. $n=1,2,3$) first and without using sum signs.
– md2perpe
Jul 17 at 11:23
Try to do the calculations in some lower dimensions (e.g. $n=1,2,3$) first and without using sum signs.
– md2perpe
Jul 17 at 11:23
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Note that if $c$ is a $k$-cell, i.e. $c : [0,1]^k to mathbb R^n$ then $c_i^rho$ is a $(k-1)$-cell, and — here is the crux — the first $i-1$ argument positions coincide with those of $c$, but the ones after the $i$th position don't; they are shifted by one step. Therefore,
$$partial c_i^rho = sum_undersetineq jj=1,ldots,k (-1)^j+[j>i]+1 (c_ij^rho 1 - c_ij^rho 0),$$
where $[j>i] = 1$ when $j>i$ and $= 0$ otherwise.
Your big sum then is symmetric in $i$ and $j$ except for the factor $(-1)^[j>i]$ which makes all terms cancel:
$$
partial^2 c
= sum_underseti neq ji,j=1,ldots,k (-1)^i+j+[j>i]+2 (c_ij^11 - c_ij^10 - c_ij^01 + c_ij^00) \
= underbracesum_underseti neq ji,j=1,ldots,k_textsymmetric in $i,j$ underbrace(-1)^[j>i]_textantisymmetric in $i,j$ underbrace (-1)^i+j+2 (c_ij^11 - c_ij^10 - c_ij^01 + c_ij^00)_textsymmetric in $i,j$
= 0$$
answered Jul 17 at 12:45
md2perpe
5,93011022
Ah, I see where my misunderstanding was. The little $[j gt i ]$ is really helpful!
– user577413
Jul 17 at 14:21
The notation is called the Iverson bracket.
– md2perpe
Jul 17 at 15:47
add a comment |Â
up vote
1
down vote
You approach is absolutely correct. Let us do it a little bit more precise.
For $rho = 0,1$ and $i = 1,...,k$ let us define maps
$$s_i,k^rho: I^k-1 to I^k, s_i,k^rho(x_1,..., x_k-1) = (x_1,...,x_i-1,rho,x_i,...,x_k-1) .$$
Then for $i le j$
$$s_i,k^rho circ s_j,k-1^sigma = s_j+1,k^sigma circ s_i,k-1^rho .$$
We have $c_i^rho = c circ s_i,k^rho$. Define $P = 1,..., k times 1,...,k-1 $, $A = (i,j) in P mid i > j $ and $B = (i,j) in P mid i le j $. We have $(i,j) in B$ if and only if $(j+1,i) in A$.
As you computed
$$
partial^2 c = sum_i=1^k sum_rho=0^1 (-1)^i+rho partial c_i^rho
\
= sum_i=1^k sum_rho=0^1 (-1)^i+rho sum_j=1 ^k-1 sum_sigma=0^1 (-1)^j+sigma (c_i^rho)_j^sigma
\
= sum_(i,j) in P sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(i,j) in B sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(i,j) in B sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_j+1,k^sigma circ s_i,k-1^rho
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(m,n) in A sum_alpha,beta = 0^1 (-1)^m+n-1+alpha+beta c circ s_m,k^alpha circ s_n,k-1^beta
\
= 0
$$
where we used $m = j+1, n = i$.
answered Jul 17 at 12:48
Paul Frost
3,703420
Thanks for the thorough answer! I can now see where the minus sign should’ve appeared.
– user577413
Jul 17 at 14:24
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note that if $c$ is a $k$-cell, i.e. $c : [0,1]^k to mathbb R^n$ then $c_i^rho$ is a $(k-1)$-cell, and — here is the crux — the first $i-1$ argument positions coincide with those of $c$, but the ones after the $i$th position don't; they are shifted by one step. Therefore,
$$partial c_i^rho = sum_undersetineq jj=1,ldots,k (-1)^j+[j>i]+1 (c_ij^rho 1 - c_ij^rho 0),$$
where $[j>i] = 1$ when $j>i$ and $= 0$ otherwise.
Your big sum then is symmetric in $i$ and $j$ except for the factor $(-1)^[j>i]$ which makes all terms cancel:
$$
partial^2 c
= sum_underseti neq ji,j=1,ldots,k (-1)^i+j+[j>i]+2 (c_ij^11 - c_ij^10 - c_ij^01 + c_ij^00) \
= underbracesum_underseti neq ji,j=1,ldots,k_textsymmetric in $i,j$ underbrace(-1)^[j>i]_textantisymmetric in $i,j$ underbrace (-1)^i+j+2 (c_ij^11 - c_ij^10 - c_ij^01 + c_ij^00)_textsymmetric in $i,j$
= 0$$
answered Jul 17 at 12:45
md2perpe
5,93011022
Ah, I see where my misunderstanding was. The little $[j gt i ]$ is really helpful!
– user577413
Jul 17 at 14:21
The notation is called the Iverson bracket.
– md2perpe
Jul 17 at 15:47
add a comment |Â
up vote
1
down vote
accepted
Note that if $c$ is a $k$-cell, i.e. $c : [0,1]^k to mathbb R^n$ then $c_i^rho$ is a $(k-1)$-cell, and — here is the crux — the first $i-1$ argument positions coincide with those of $c$, but the ones after the $i$th position don't; they are shifted by one step. Therefore,
$$partial c_i^rho = sum_undersetineq jj=1,ldots,k (-1)^j+[j>i]+1 (c_ij^rho 1 - c_ij^rho 0),$$
where $[j>i] = 1$ when $j>i$ and $= 0$ otherwise.
Your big sum then is symmetric in $i$ and $j$ except for the factor $(-1)^[j>i]$ which makes all terms cancel:
$$
partial^2 c
= sum_underseti neq ji,j=1,ldots,k (-1)^i+j+[j>i]+2 (c_ij^11 - c_ij^10 - c_ij^01 + c_ij^00) \
= underbracesum_underseti neq ji,j=1,ldots,k_textsymmetric in $i,j$ underbrace(-1)^[j>i]_textantisymmetric in $i,j$ underbrace (-1)^i+j+2 (c_ij^11 - c_ij^10 - c_ij^01 + c_ij^00)_textsymmetric in $i,j$
= 0$$
answered Jul 17 at 12:45
md2perpe
5,93011022
Ah, I see where my misunderstanding was. The little $[j gt i ]$ is really helpful!
– user577413
Jul 17 at 14:21
The notation is called the Iverson bracket.
– md2perpe
Jul 17 at 15:47
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note that if $c$ is a $k$-cell, i.e. $c : [0,1]^k to mathbb R^n$ then $c_i^rho$ is a $(k-1)$-cell, and — here is the crux — the first $i-1$ argument positions coincide with those of $c$, but the ones after the $i$th position don't; they are shifted by one step. Therefore,
$$partial c_i^rho = sum_undersetineq jj=1,ldots,k (-1)^j+[j>i]+1 (c_ij^rho 1 - c_ij^rho 0),$$
where $[j>i] = 1$ when $j>i$ and $= 0$ otherwise.
Your big sum then is symmetric in $i$ and $j$ except for the factor $(-1)^[j>i]$ which makes all terms cancel:
$$
partial^2 c
= sum_underseti neq ji,j=1,ldots,k (-1)^i+j+[j>i]+2 (c_ij^11 - c_ij^10 - c_ij^01 + c_ij^00) \
= underbracesum_underseti neq ji,j=1,ldots,k_textsymmetric in $i,j$ underbrace(-1)^[j>i]_textantisymmetric in $i,j$ underbrace (-1)^i+j+2 (c_ij^11 - c_ij^10 - c_ij^01 + c_ij^00)_textsymmetric in $i,j$
= 0$$
answered Jul 17 at 12:45
md2perpe
5,93011022
Note that if $c$ is a $k$-cell, i.e. $c : [0,1]^k to mathbb R^n$ then $c_i^rho$ is a $(k-1)$-cell, and — here is the crux — the first $i-1$ argument positions coincide with those of $c$, but the ones after the $i$th position don't; they are shifted by one step. Therefore,
$$partial c_i^rho = sum_undersetineq jj=1,ldots,k (-1)^j+[j>i]+1 (c_ij^rho 1 - c_ij^rho 0),$$
where $[j>i] = 1$ when $j>i$ and $= 0$ otherwise.
Your big sum then is symmetric in $i$ and $j$ except for the factor $(-1)^[j>i]$ which makes all terms cancel:
$$
partial^2 c
= sum_underseti neq ji,j=1,ldots,k (-1)^i+j+[j>i]+2 (c_ij^11 - c_ij^10 - c_ij^01 + c_ij^00) \
= underbracesum_underseti neq ji,j=1,ldots,k_textsymmetric in $i,j$ underbrace(-1)^[j>i]_textantisymmetric in $i,j$ underbrace (-1)^i+j+2 (c_ij^11 - c_ij^10 - c_ij^01 + c_ij^00)_textsymmetric in $i,j$
= 0$$
answered Jul 17 at 12:45
md2perpe
5,93011022
answered Jul 17 at 12:45
md2perpe
5,93011022
answered Jul 17 at 12:45
md2perpe
5,93011022
answered Jul 17 at 12:45
md2perpe
5,93011022
5,93011022
Ah, I see where my misunderstanding was. The little $[j gt i ]$ is really helpful!
– user577413
Jul 17 at 14:21
The notation is called the Iverson bracket.
– md2perpe
Jul 17 at 15:47
add a comment |Â
Ah, I see where my misunderstanding was. The little $[j gt i ]$ is really helpful!
– user577413
Jul 17 at 14:21
The notation is called the Iverson bracket.
– md2perpe
Jul 17 at 15:47
Ah, I see where my misunderstanding was. The little $[j gt i ]$ is really helpful!
– user577413
Jul 17 at 14:21
Ah, I see where my misunderstanding was. The little $[j gt i ]$ is really helpful!
– user577413
Jul 17 at 14:21
The notation is called the Iverson bracket.
– md2perpe
Jul 17 at 15:47
The notation is called the Iverson bracket.
– md2perpe
Jul 17 at 15:47
add a comment |Â
up vote
1
down vote
You approach is absolutely correct. Let us do it a little bit more precise.
For $rho = 0,1$ and $i = 1,...,k$ let us define maps
$$s_i,k^rho: I^k-1 to I^k, s_i,k^rho(x_1,..., x_k-1) = (x_1,...,x_i-1,rho,x_i,...,x_k-1) .$$
Then for $i le j$
$$s_i,k^rho circ s_j,k-1^sigma = s_j+1,k^sigma circ s_i,k-1^rho .$$
We have $c_i^rho = c circ s_i,k^rho$. Define $P = 1,..., k times 1,...,k-1 $, $A = (i,j) in P mid i > j $ and $B = (i,j) in P mid i le j $. We have $(i,j) in B$ if and only if $(j+1,i) in A$.
As you computed
$$
partial^2 c = sum_i=1^k sum_rho=0^1 (-1)^i+rho partial c_i^rho
\
= sum_i=1^k sum_rho=0^1 (-1)^i+rho sum_j=1 ^k-1 sum_sigma=0^1 (-1)^j+sigma (c_i^rho)_j^sigma
\
= sum_(i,j) in P sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(i,j) in B sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(i,j) in B sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_j+1,k^sigma circ s_i,k-1^rho
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(m,n) in A sum_alpha,beta = 0^1 (-1)^m+n-1+alpha+beta c circ s_m,k^alpha circ s_n,k-1^beta
\
= 0
$$
where we used $m = j+1, n = i$.
answered Jul 17 at 12:48
Paul Frost
3,703420
Thanks for the thorough answer! I can now see where the minus sign should’ve appeared.
– user577413
Jul 17 at 14:24
add a comment |Â
up vote
1
down vote
You approach is absolutely correct. Let us do it a little bit more precise.
For $rho = 0,1$ and $i = 1,...,k$ let us define maps
$$s_i,k^rho: I^k-1 to I^k, s_i,k^rho(x_1,..., x_k-1) = (x_1,...,x_i-1,rho,x_i,...,x_k-1) .$$
Then for $i le j$
$$s_i,k^rho circ s_j,k-1^sigma = s_j+1,k^sigma circ s_i,k-1^rho .$$
We have $c_i^rho = c circ s_i,k^rho$. Define $P = 1,..., k times 1,...,k-1 $, $A = (i,j) in P mid i > j $ and $B = (i,j) in P mid i le j $. We have $(i,j) in B$ if and only if $(j+1,i) in A$.
As you computed
$$
partial^2 c = sum_i=1^k sum_rho=0^1 (-1)^i+rho partial c_i^rho
\
= sum_i=1^k sum_rho=0^1 (-1)^i+rho sum_j=1 ^k-1 sum_sigma=0^1 (-1)^j+sigma (c_i^rho)_j^sigma
\
= sum_(i,j) in P sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(i,j) in B sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(i,j) in B sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_j+1,k^sigma circ s_i,k-1^rho
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(m,n) in A sum_alpha,beta = 0^1 (-1)^m+n-1+alpha+beta c circ s_m,k^alpha circ s_n,k-1^beta
\
= 0
$$
where we used $m = j+1, n = i$.
answered Jul 17 at 12:48
Paul Frost
3,703420
Thanks for the thorough answer! I can now see where the minus sign should’ve appeared.
– user577413
Jul 17 at 14:24
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You approach is absolutely correct. Let us do it a little bit more precise.
For $rho = 0,1$ and $i = 1,...,k$ let us define maps
$$s_i,k^rho: I^k-1 to I^k, s_i,k^rho(x_1,..., x_k-1) = (x_1,...,x_i-1,rho,x_i,...,x_k-1) .$$
Then for $i le j$
$$s_i,k^rho circ s_j,k-1^sigma = s_j+1,k^sigma circ s_i,k-1^rho .$$
We have $c_i^rho = c circ s_i,k^rho$. Define $P = 1,..., k times 1,...,k-1 $, $A = (i,j) in P mid i > j $ and $B = (i,j) in P mid i le j $. We have $(i,j) in B$ if and only if $(j+1,i) in A$.
As you computed
$$
partial^2 c = sum_i=1^k sum_rho=0^1 (-1)^i+rho partial c_i^rho
\
= sum_i=1^k sum_rho=0^1 (-1)^i+rho sum_j=1 ^k-1 sum_sigma=0^1 (-1)^j+sigma (c_i^rho)_j^sigma
\
= sum_(i,j) in P sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(i,j) in B sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(i,j) in B sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_j+1,k^sigma circ s_i,k-1^rho
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(m,n) in A sum_alpha,beta = 0^1 (-1)^m+n-1+alpha+beta c circ s_m,k^alpha circ s_n,k-1^beta
\
= 0
$$
where we used $m = j+1, n = i$.
answered Jul 17 at 12:48
Paul Frost
3,703420
You approach is absolutely correct. Let us do it a little bit more precise.
For $rho = 0,1$ and $i = 1,...,k$ let us define maps
$$s_i,k^rho: I^k-1 to I^k, s_i,k^rho(x_1,..., x_k-1) = (x_1,...,x_i-1,rho,x_i,...,x_k-1) .$$
Then for $i le j$
$$s_i,k^rho circ s_j,k-1^sigma = s_j+1,k^sigma circ s_i,k-1^rho .$$
We have $c_i^rho = c circ s_i,k^rho$. Define $P = 1,..., k times 1,...,k-1 $, $A = (i,j) in P mid i > j $ and $B = (i,j) in P mid i le j $. We have $(i,j) in B$ if and only if $(j+1,i) in A$.
As you computed
$$
partial^2 c = sum_i=1^k sum_rho=0^1 (-1)^i+rho partial c_i^rho
\
= sum_i=1^k sum_rho=0^1 (-1)^i+rho sum_j=1 ^k-1 sum_sigma=0^1 (-1)^j+sigma (c_i^rho)_j^sigma
\
= sum_(i,j) in P sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(i,j) in B sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(i,j) in B sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_j+1,k^sigma circ s_i,k-1^rho
\
= sum_(i,j) in A sum_rho,sigma = 0^1 (-1)^i+j+rho+sigma c circ s_i,k^rho circ s_j,k-1^sigma + sum_(m,n) in A sum_alpha,beta = 0^1 (-1)^m+n-1+alpha+beta c circ s_m,k^alpha circ s_n,k-1^beta
\
= 0
$$
where we used $m = j+1, n = i$.
answered Jul 17 at 12:48
Paul Frost
3,703420
answered Jul 17 at 12:48
Paul Frost
3,703420
answered Jul 17 at 12:48
Paul Frost
3,703420
answered Jul 17 at 12:48
Paul Frost
3,703420
3,703420
Thanks for the thorough answer! I can now see where the minus sign should’ve appeared.
– user577413
Jul 17 at 14:24
add a comment |Â
Thanks for the thorough answer! I can now see where the minus sign should’ve appeared.
– user577413
Jul 17 at 14:24
Thanks for the thorough answer! I can now see where the minus sign should’ve appeared.
– user577413
Jul 17 at 14:24
Thanks for the thorough answer! I can now see where the minus sign should’ve appeared.
– user577413
Jul 17 at 14:24
add a comment |Â
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Try to do the calculations in some lower dimensions (e.g. $n=1,2,3$) first and without using sum signs.
– md2perpe
Jul 17 at 11:23