Projective Resolution of exterior algebra as a module over divided polynomial algebra The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Projective dimenson of tensor productDefinition for Ext and TorResolution of an algebraExample of module/ring with long projective resolutionWhy is the bar complex free?Calculate $textExt_mathbbZ^1(mathbbQ/mathbbZ, mathbbZ) $?Why does group homology do not depend on the coefficient ring?Why can a projective resolution of $A$ be used to calculate $Ext_R^n(A,B)$?Divided power algebra is artinian as a module over the polynomial ringCalculation of Tor for module with trivial action
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Projective Resolution of exterior algebra as a module over divided polynomial algebra
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Projective dimenson of tensor productDefinition for Ext and TorResolution of an algebraExample of module/ring with long projective resolutionWhy is the bar complex free?Calculate $textExt_mathbbZ^1(mathbbQ/mathbbZ, mathbbZ) $?Why does group homology do not depend on the coefficient ring?Why can a projective resolution of $A$ be used to calculate $Ext_R^n(A,B)$?Divided power algebra is artinian as a module over the polynomial ringCalculation of Tor for module with trivial action
$begingroup$
Let $Lambda_mathbbZ[x]$ be an exterior algebra on one generator with $|x|=n$, let $Gamma_mathbbZ[x]$ be a divided polynomial algebra with $|x_k|= kn$, and suppose that $Lambda_mathbbZ[x]$ is a module over $Gamma_mathbbZ[x]$.
The multiplication on $Gamma_mathbbZ[x]$ is given by
$$
x_kx_l = k+l choose k x_k+l.
$$
Question: Calculate a projective resolution for $Lambda_mathbbZ[x]$ as a module over $Gamma_mathbbZ[x]$.
Attempt
I'm attempting to answer this without any overly technical machinery (bar construction/Kozul complex etc). I will denote $mathbbZ_x_k$ a copy of the integers generated by $x_k$.
Any help with this would be greatly appreciated. I am only interested in the case up to $mathbbZ_x_4$ as it is all I need for the calculation purposes.
Any help would be greatly appreciated.
homological-algebra projective-module graded-modules graded-algebras
asked Mar 25 at 9:40
CharlesCharles
20818
$endgroup$
add a comment |
$begingroup$
Let $Lambda_mathbbZ[x]$ be an exterior algebra on one generator with $|x|=n$, let $Gamma_mathbbZ[x]$ be a divided polynomial algebra with $|x_k|= kn$, and suppose that $Lambda_mathbbZ[x]$ is a module over $Gamma_mathbbZ[x]$.
The multiplication on $Gamma_mathbbZ[x]$ is given by
$$
x_kx_l = k+l choose k x_k+l.
$$
Question: Calculate a projective resolution for $Lambda_mathbbZ[x]$ as a module over $Gamma_mathbbZ[x]$.
Attempt
I'm attempting to answer this without any overly technical machinery (bar construction/Kozul complex etc). I will denote $mathbbZ_x_k$ a copy of the integers generated by $x_k$.
Any help with this would be greatly appreciated. I am only interested in the case up to $mathbbZ_x_4$ as it is all I need for the calculation purposes.
Any help would be greatly appreciated.
homological-algebra projective-module graded-modules graded-algebras
asked Mar 25 at 9:40
CharlesCharles
20818
$endgroup$
add a comment |
$begingroup$
Let $Lambda_mathbbZ[x]$ be an exterior algebra on one generator with $|x|=n$, let $Gamma_mathbbZ[x]$ be a divided polynomial algebra with $|x_k|= kn$, and suppose that $Lambda_mathbbZ[x]$ is a module over $Gamma_mathbbZ[x]$.
The multiplication on $Gamma_mathbbZ[x]$ is given by
$$
x_kx_l = k+l choose k x_k+l.
$$
Question: Calculate a projective resolution for $Lambda_mathbbZ[x]$ as a module over $Gamma_mathbbZ[x]$.
Attempt
I'm attempting to answer this without any overly technical machinery (bar construction/Kozul complex etc). I will denote $mathbbZ_x_k$ a copy of the integers generated by $x_k$.
Any help with this would be greatly appreciated. I am only interested in the case up to $mathbbZ_x_4$ as it is all I need for the calculation purposes.
Any help would be greatly appreciated.
homological-algebra projective-module graded-modules graded-algebras
asked Mar 25 at 9:40
CharlesCharles
20818
$endgroup$
Let $Lambda_mathbbZ[x]$ be an exterior algebra on one generator with $|x|=n$, let $Gamma_mathbbZ[x]$ be a divided polynomial algebra with $|x_k|= kn$, and suppose that $Lambda_mathbbZ[x]$ is a module over $Gamma_mathbbZ[x]$.
The multiplication on $Gamma_mathbbZ[x]$ is given by
$$
x_kx_l = k+l choose k x_k+l.
$$
Question: Calculate a projective resolution for $Lambda_mathbbZ[x]$ as a module over $Gamma_mathbbZ[x]$.
Attempt
I'm attempting to answer this without any overly technical machinery (bar construction/Kozul complex etc). I will denote $mathbbZ_x_k$ a copy of the integers generated by $x_k$.
Any help with this would be greatly appreciated. I am only interested in the case up to $mathbbZ_x_4$ as it is all I need for the calculation purposes.
Any help would be greatly appreciated.
homological-algebra projective-module graded-modules graded-algebras
homological-algebra projective-module graded-modules graded-algebras
asked Mar 25 at 9:40
CharlesCharles
20818
asked Mar 25 at 9:40
CharlesCharles
20818
edited Mar 25 at 18:33
Charles
asked Mar 25 at 9:40
CharlesCharles
20818
asked Mar 25 at 9:40
CharlesCharles
20818
asked Mar 25 at 9:40
CharlesCharles
20818
20818
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can continue computing a minimal free resolution as you have done. If I haven't made a mistake, the answer, up through degree $4n$, is $Gamma_mathbbZ[x] cdot alpha_0, beta_2, gamma_3, delta_4, epsilon_3, zeta_4, eta_4, theta_4$ where the degree of each basis element is $n$ times the subscript. The differentials are:
beginalign*
d(alpha_0) &= 1 \
d(beta_2) &= x_2 alpha_0 \
d(gamma_3) &= x_3 alpha_0 \
d(delta_4) &= x_4 alpha_0 \
d(epsilon_3) &= 3gamma_3 - x_1 beta_2 \
d(zeta_4) &= 6delta_4 - x_2 beta_2 \
d(eta_4) &= 4delta_4 - x_1 gamma_3 \
d(theta_4) &= x_1 epsilon_3 + 2zeta_4 - 3eta_4.
endalign*
answered Mar 25 at 23:42
JHFJHF
4,9961026
$endgroup$
$begingroup$
What does the notation $Gamma_mathbbZ[x] cdot alpha_0, beta_2, gamma_3, delta_4, epsilon_3, zeta_4, eta_4, theta_4$ mean? Would it be possible for you to display this in a similar fashion to how I have above
$endgroup$
– Charles
Mar 26 at 9:48
$begingroup$
The notation just means a free $Gamma_mathbbZ[x]$ module with generators $alpha_0, beta_2, ldots, theta_4$ of the appropriate degrees. Unfortunately, my knowledge of MathJax is not good enough to TeX up a diagram. If it helps, $alpha_0$ has homological degree $0$, $beta_2, gamma_3, delta_4$ has degree $1$, $epsilon_3, zeta_4, eta_4$ has degree $2$, and $theta_4$ has degree $3$.
$endgroup$
– JHF
Mar 26 at 18:03
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can continue computing a minimal free resolution as you have done. If I haven't made a mistake, the answer, up through degree $4n$, is $Gamma_mathbbZ[x] cdot alpha_0, beta_2, gamma_3, delta_4, epsilon_3, zeta_4, eta_4, theta_4$ where the degree of each basis element is $n$ times the subscript. The differentials are:
beginalign*
d(alpha_0) &= 1 \
d(beta_2) &= x_2 alpha_0 \
d(gamma_3) &= x_3 alpha_0 \
d(delta_4) &= x_4 alpha_0 \
d(epsilon_3) &= 3gamma_3 - x_1 beta_2 \
d(zeta_4) &= 6delta_4 - x_2 beta_2 \
d(eta_4) &= 4delta_4 - x_1 gamma_3 \
d(theta_4) &= x_1 epsilon_3 + 2zeta_4 - 3eta_4.
endalign*
answered Mar 25 at 23:42
JHFJHF
4,9961026
$endgroup$
$begingroup$
What does the notation $Gamma_mathbbZ[x] cdot alpha_0, beta_2, gamma_3, delta_4, epsilon_3, zeta_4, eta_4, theta_4$ mean? Would it be possible for you to display this in a similar fashion to how I have above
$endgroup$
– Charles
Mar 26 at 9:48
$begingroup$
The notation just means a free $Gamma_mathbbZ[x]$ module with generators $alpha_0, beta_2, ldots, theta_4$ of the appropriate degrees. Unfortunately, my knowledge of MathJax is not good enough to TeX up a diagram. If it helps, $alpha_0$ has homological degree $0$, $beta_2, gamma_3, delta_4$ has degree $1$, $epsilon_3, zeta_4, eta_4$ has degree $2$, and $theta_4$ has degree $3$.
$endgroup$
– JHF
Mar 26 at 18:03
add a comment |
$begingroup$
We can continue computing a minimal free resolution as you have done. If I haven't made a mistake, the answer, up through degree $4n$, is $Gamma_mathbbZ[x] cdot alpha_0, beta_2, gamma_3, delta_4, epsilon_3, zeta_4, eta_4, theta_4$ where the degree of each basis element is $n$ times the subscript. The differentials are:
beginalign*
d(alpha_0) &= 1 \
d(beta_2) &= x_2 alpha_0 \
d(gamma_3) &= x_3 alpha_0 \
d(delta_4) &= x_4 alpha_0 \
d(epsilon_3) &= 3gamma_3 - x_1 beta_2 \
d(zeta_4) &= 6delta_4 - x_2 beta_2 \
d(eta_4) &= 4delta_4 - x_1 gamma_3 \
d(theta_4) &= x_1 epsilon_3 + 2zeta_4 - 3eta_4.
endalign*
answered Mar 25 at 23:42
JHFJHF
4,9961026
$endgroup$
$begingroup$
What does the notation $Gamma_mathbbZ[x] cdot alpha_0, beta_2, gamma_3, delta_4, epsilon_3, zeta_4, eta_4, theta_4$ mean? Would it be possible for you to display this in a similar fashion to how I have above
$endgroup$
– Charles
Mar 26 at 9:48
$begingroup$
The notation just means a free $Gamma_mathbbZ[x]$ module with generators $alpha_0, beta_2, ldots, theta_4$ of the appropriate degrees. Unfortunately, my knowledge of MathJax is not good enough to TeX up a diagram. If it helps, $alpha_0$ has homological degree $0$, $beta_2, gamma_3, delta_4$ has degree $1$, $epsilon_3, zeta_4, eta_4$ has degree $2$, and $theta_4$ has degree $3$.
$endgroup$
– JHF
Mar 26 at 18:03
add a comment |
$begingroup$
We can continue computing a minimal free resolution as you have done. If I haven't made a mistake, the answer, up through degree $4n$, is $Gamma_mathbbZ[x] cdot alpha_0, beta_2, gamma_3, delta_4, epsilon_3, zeta_4, eta_4, theta_4$ where the degree of each basis element is $n$ times the subscript. The differentials are:
beginalign*
d(alpha_0) &= 1 \
d(beta_2) &= x_2 alpha_0 \
d(gamma_3) &= x_3 alpha_0 \
d(delta_4) &= x_4 alpha_0 \
d(epsilon_3) &= 3gamma_3 - x_1 beta_2 \
d(zeta_4) &= 6delta_4 - x_2 beta_2 \
d(eta_4) &= 4delta_4 - x_1 gamma_3 \
d(theta_4) &= x_1 epsilon_3 + 2zeta_4 - 3eta_4.
endalign*
answered Mar 25 at 23:42
JHFJHF
4,9961026
$endgroup$
We can continue computing a minimal free resolution as you have done. If I haven't made a mistake, the answer, up through degree $4n$, is $Gamma_mathbbZ[x] cdot alpha_0, beta_2, gamma_3, delta_4, epsilon_3, zeta_4, eta_4, theta_4$ where the degree of each basis element is $n$ times the subscript. The differentials are:
beginalign*
d(alpha_0) &= 1 \
d(beta_2) &= x_2 alpha_0 \
d(gamma_3) &= x_3 alpha_0 \
d(delta_4) &= x_4 alpha_0 \
d(epsilon_3) &= 3gamma_3 - x_1 beta_2 \
d(zeta_4) &= 6delta_4 - x_2 beta_2 \
d(eta_4) &= 4delta_4 - x_1 gamma_3 \
d(theta_4) &= x_1 epsilon_3 + 2zeta_4 - 3eta_4.
endalign*
answered Mar 25 at 23:42
JHFJHF
4,9961026
answered Mar 25 at 23:42
JHFJHF
4,9961026
answered Mar 25 at 23:42
JHFJHF
4,9961026
answered Mar 25 at 23:42
JHFJHF
4,9961026
4,9961026
$begingroup$
What does the notation $Gamma_mathbbZ[x] cdot alpha_0, beta_2, gamma_3, delta_4, epsilon_3, zeta_4, eta_4, theta_4$ mean? Would it be possible for you to display this in a similar fashion to how I have above
$endgroup$
– Charles
Mar 26 at 9:48
$begingroup$
The notation just means a free $Gamma_mathbbZ[x]$ module with generators $alpha_0, beta_2, ldots, theta_4$ of the appropriate degrees. Unfortunately, my knowledge of MathJax is not good enough to TeX up a diagram. If it helps, $alpha_0$ has homological degree $0$, $beta_2, gamma_3, delta_4$ has degree $1$, $epsilon_3, zeta_4, eta_4$ has degree $2$, and $theta_4$ has degree $3$.
$endgroup$
– JHF
Mar 26 at 18:03
add a comment |
$begingroup$
What does the notation $Gamma_mathbbZ[x] cdot alpha_0, beta_2, gamma_3, delta_4, epsilon_3, zeta_4, eta_4, theta_4$ mean? Would it be possible for you to display this in a similar fashion to how I have above
$endgroup$
– Charles
Mar 26 at 9:48
$begingroup$
The notation just means a free $Gamma_mathbbZ[x]$ module with generators $alpha_0, beta_2, ldots, theta_4$ of the appropriate degrees. Unfortunately, my knowledge of MathJax is not good enough to TeX up a diagram. If it helps, $alpha_0$ has homological degree $0$, $beta_2, gamma_3, delta_4$ has degree $1$, $epsilon_3, zeta_4, eta_4$ has degree $2$, and $theta_4$ has degree $3$.
$endgroup$
– JHF
Mar 26 at 18:03
$begingroup$
What does the notation $Gamma_mathbbZ[x] cdot alpha_0, beta_2, gamma_3, delta_4, epsilon_3, zeta_4, eta_4, theta_4$ mean? Would it be possible for you to display this in a similar fashion to how I have above
$endgroup$
– Charles
Mar 26 at 9:48
$begingroup$
What does the notation $Gamma_mathbbZ[x] cdot alpha_0, beta_2, gamma_3, delta_4, epsilon_3, zeta_4, eta_4, theta_4$ mean? Would it be possible for you to display this in a similar fashion to how I have above
$endgroup$
– Charles
Mar 26 at 9:48
$begingroup$
The notation just means a free $Gamma_mathbbZ[x]$ module with generators $alpha_0, beta_2, ldots, theta_4$ of the appropriate degrees. Unfortunately, my knowledge of MathJax is not good enough to TeX up a diagram. If it helps, $alpha_0$ has homological degree $0$, $beta_2, gamma_3, delta_4$ has degree $1$, $epsilon_3, zeta_4, eta_4$ has degree $2$, and $theta_4$ has degree $3$.
$endgroup$
– JHF
Mar 26 at 18:03
$begingroup$
The notation just means a free $Gamma_mathbbZ[x]$ module with generators $alpha_0, beta_2, ldots, theta_4$ of the appropriate degrees. Unfortunately, my knowledge of MathJax is not good enough to TeX up a diagram. If it helps, $alpha_0$ has homological degree $0$, $beta_2, gamma_3, delta_4$ has degree $1$, $epsilon_3, zeta_4, eta_4$ has degree $2$, and $theta_4$ has degree $3$.
$endgroup$
– JHF
Mar 26 at 18:03
add a comment |
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