Fdtxjr

Who or what is the being for whom Being is a question for Heidegger?

Relations between two reciprocal partial derivatives?

Is every episode of "Where are my Pants?" identical?

What's the point in a preamp?

Why is the object placed in the middle of the sentence here?

Reference for the teaching of not-self

Single author papers against my advisor's will?

How long does the line of fire that you can create as an action using the Investiture of Flame spell last?

Slither Like a Snake

How does ice melt when immersed in water?

How many people can fit inside Mordenkainen's Magnificent Mansion?

Why does the Event Horizon Telescope (EHT) not include telescopes from Africa, Asia or Australia?

Are my PIs rude or am I just being too sensitive?

The following signatures were invalid: EXPKEYSIG 1397BC53640DB551

How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time

What information about me do stores get via my credit card?

Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?

How can I define good in a religion that claims no moral authority?

How to pronounce 1ターン?

How to copy the contents of all files with a certain name into a new file?

Is it ethical to upload a automatically generated paper to a non peer-reviewed site as part of a larger research?

Was credit for the black hole image misattributed?

Working through the single responsibility principle (SRP) in Python when calls are expensive

What was the last x86 CPU that did not have the x87 floating-point unit built in?

Projective Resolution of exterior algebra as a module over divided polynomial algebra

1

0

$begingroup$

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

share|cite|improve this question

edited Mar 25 at 18:33

Charles

asked Mar 25 at 9:40

CharlesCharles

20818

$endgroup$

add a comment | 

1

0

$begingroup$

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

share|cite|improve this question

edited Mar 25 at 18:33

Charles

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

share|cite|improve this question

edited Mar 25 at 18:33

Charles

asked Mar 25 at 9:40

CharlesCharles

20818

$endgroup$

share|cite|improve this question

edited Mar 25 at 18:33

Charles

asked Mar 25 at 9:40

CharlesCharles

20818

share|cite|improve this question

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

1 Answer
1

active

oldest

votes

0

$begingroup$

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*

share|cite|improve this answer

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 | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3161569%2fprojective-resolution-of-exterior-algebra-as-a-module-over-divided-polynomial-al%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

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*

share|cite|improve this answer

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 | 

0

$begingroup$

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*

share|cite|improve this answer

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*

share|cite|improve this answer

answered Mar 25 at 23:42

JHFJHF

4,9961026

$endgroup$

share|cite|improve this answer

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