Acyclic Chain Complex
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I’m a little confused, for a chain to be acyclic, all Betti numbers must be zero. For a Betti number $beta$
$beta_i=dim(Z_i)-dim(B_i)$ where $Z_i=ker(partial_i)$ and $B_i$=im$(partial_i{+}_1))$
I have $beta_1=dim(Z_1)-dim(B_1)=2-3$, am I going about this the right way?
linear-algebra topological-data-analysis
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
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add a comment |
$begingroup$
I’m a little confused, for a chain to be acyclic, all Betti numbers must be zero. For a Betti number $beta$
$beta_i=dim(Z_i)-dim(B_i)$ where $Z_i=ker(partial_i)$ and $B_i$=im$(partial_i{+}_1))$
I have $beta_1=dim(Z_1)-dim(B_1)=2-3$, am I going about this the right way?
linear-algebra topological-data-analysis
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
$endgroup$
$begingroup$
Isn't the image of the arrow $mathbb{R}^2tomathbb{R}^3$ $2$-dimensional, not $3$-dimensional, if I'm interpreting your indexing correctly?
$endgroup$
– BWW
Dec 4 '18 at 1:03
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Wouldn’t $B_1$=im$(T_2)=span{[1 -1 1]}$?, and that would be 3 dimensional?
$endgroup$
– LexyFidds
Dec 4 '18 at 1:09
$begingroup$
Aaaahhh you’re right, I think I’ve been indexing incorrectly
$endgroup$
– LexyFidds
Dec 4 '18 at 1:12
add a comment |
$begingroup$
I’m a little confused, for a chain to be acyclic, all Betti numbers must be zero. For a Betti number $beta$
$beta_i=dim(Z_i)-dim(B_i)$ where $Z_i=ker(partial_i)$ and $B_i$=im$(partial_i{+}_1))$
I have $beta_1=dim(Z_1)-dim(B_1)=2-3$, am I going about this the right way?
linear-algebra topological-data-analysis
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
$endgroup$
I’m a little confused, for a chain to be acyclic, all Betti numbers must be zero. For a Betti number $beta$
$beta_i=dim(Z_i)-dim(B_i)$ where $Z_i=ker(partial_i)$ and $B_i$=im$(partial_i{+}_1))$
I have $beta_1=dim(Z_1)-dim(B_1)=2-3$, am I going about this the right way?
linear-algebra topological-data-analysis
linear-algebra topological-data-analysis
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
edited Dec 4 '18 at 0:58
LexyFidds
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
asked Dec 4 '18 at 0:54
LexyFidds LexyFidds
226
226
$begingroup$
Isn't the image of the arrow $mathbb{R}^2tomathbb{R}^3$ $2$-dimensional, not $3$-dimensional, if I'm interpreting your indexing correctly?
$endgroup$
– BWW
Dec 4 '18 at 1:03
$begingroup$
Wouldn’t $B_1$=im$(T_2)=span{[1 -1 1]}$?, and that would be 3 dimensional?
$endgroup$
– LexyFidds
Dec 4 '18 at 1:09
$begingroup$
Aaaahhh you’re right, I think I’ve been indexing incorrectly
$endgroup$
– LexyFidds
Dec 4 '18 at 1:12
add a comment |
$begingroup$
Isn't the image of the arrow $mathbb{R}^2tomathbb{R}^3$ $2$-dimensional, not $3$-dimensional, if I'm interpreting your indexing correctly?
$endgroup$
– BWW
Dec 4 '18 at 1:03
$begingroup$
Wouldn’t $B_1$=im$(T_2)=span{[1 -1 1]}$?, and that would be 3 dimensional?
$endgroup$
– LexyFidds
Dec 4 '18 at 1:09
$begingroup$
Aaaahhh you’re right, I think I’ve been indexing incorrectly
$endgroup$
– LexyFidds
Dec 4 '18 at 1:12
$begingroup$
Isn't the image of the arrow $mathbb{R}^2tomathbb{R}^3$ $2$-dimensional, not $3$-dimensional, if I'm interpreting your indexing correctly?
$endgroup$
– BWW
Dec 4 '18 at 1:03
$begingroup$
Isn't the image of the arrow $mathbb{R}^2tomathbb{R}^3$ $2$-dimensional, not $3$-dimensional, if I'm interpreting your indexing correctly?
$endgroup$
– BWW
Dec 4 '18 at 1:03
$begingroup$
Wouldn’t $B_1$=im$(T_2)=span{[1 -1 1]}$?, and that would be 3 dimensional?
$endgroup$
– LexyFidds
Dec 4 '18 at 1:09
$begingroup$
Wouldn’t $B_1$=im$(T_2)=span{[1 -1 1]}$?, and that would be 3 dimensional?
$endgroup$
– LexyFidds
Dec 4 '18 at 1:09
$begingroup$
Aaaahhh you’re right, I think I’ve been indexing incorrectly
$endgroup$
– LexyFidds
Dec 4 '18 at 1:12
$begingroup$
Aaaahhh you’re right, I think I’ve been indexing incorrectly
$endgroup$
– LexyFidds
Dec 4 '18 at 1:12
add a comment |
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$begingroup$
Isn't the image of the arrow $mathbb{R}^2tomathbb{R}^3$ $2$-dimensional, not $3$-dimensional, if I'm interpreting your indexing correctly?
$endgroup$
– BWW
Dec 4 '18 at 1:03
$begingroup$
Wouldn’t $B_1$=im$(T_2)=span{[1 -1 1]}$?, and that would be 3 dimensional?
$endgroup$
– LexyFidds
Dec 4 '18 at 1:09
$begingroup$
Aaaahhh you’re right, I think I’ve been indexing incorrectly
$endgroup$
– LexyFidds
Dec 4 '18 at 1:12