Mixing
Problem with universal enveloping algebra generated by a single element (Jacobson)
$begingroup$
Jacobson's book on "Lie algebras" has the following definition of enveloping algebra generated by a subset (Definition 2, Chap II) : Start with an unital associative algebra $A$ (over a field $F$) and $S subset A$. The enveloping algebra $S^{ast}$ is simply the associative subalgebra of $A$ containing $1_A$ generated by $S$. The way I see it as $S^{ast} = sum_{n geq 0} S^n$ where $S^n$ is the $F$-submodule generated by the set ${ s_1 dotsc s_n ~:~ s_1, dotsc, s_n in S }$ of $n$-fold monomials (of course $S^{0} := F$). Since we require $1_A in S^{ast}$, we need $n=0$ in the sum.
Now while describing the subsequent properties, it says for $w in A$ we have ${ w }^{ast}$ is the algebra of polynomials in $w$ with constant term $0$. How can we have the constant terms $0$ if I need $1_A in { w }^{ast}$?
lie-algebras
asked Jan 20 at 17:24
SiddharthaSiddhartha
416
$endgroup$
add a comment |
$begingroup$
Jacobson's book on "Lie algebras" has the following definition of enveloping algebra generated by a subset (Definition 2, Chap II) : Start with an unital associative algebra $A$ (over a field $F$) and $S subset A$. The enveloping algebra $S^{ast}$ is simply the associative subalgebra of $A$ containing $1_A$ generated by $S$. The way I see it as $S^{ast} = sum_{n geq 0} S^n$ where $S^n$ is the $F$-submodule generated by the set ${ s_1 dotsc s_n ~:~ s_1, dotsc, s_n in S }$ of $n$-fold monomials (of course $S^{0} := F$). Since we require $1_A in S^{ast}$, we need $n=0$ in the sum.
Now while describing the subsequent properties, it says for $w in A$ we have ${ w }^{ast}$ is the algebra of polynomials in $w$ with constant term $0$. How can we have the constant terms $0$ if I need $1_A in { w }^{ast}$?
lie-algebras
asked Jan 20 at 17:24
SiddharthaSiddhartha
416
$endgroup$
$begingroup$
I think(?) I found the problem : while discussing the enveloping sub-algebra (containing 1) generated by $S$ the book uses the notation $S^{ast}(S^{dagger})$, elsewhere it simply uses $S^{ast}$. Maybe this is the difference between whether to take $n=0$ or not in the sum of the definition. There is no a priori clarification of these, and so I am not sure.
$endgroup$
– Siddhartha
Jan 21 at 14:12
$begingroup$
This has nothing to do with Lie algebras. It's only about associative algebras. You have added "universal" in the title but it's not called "universal" in Jacobson. Anyway, what Jacobson calls "enveloping algebra" should have been called "enveloping subalgebra", and is now rather referred to as "unital subalgebra generated by $S$".
$endgroup$
– YCor
Jan 27 at 4:13
add a comment |
$begingroup$
Jacobson's book on "Lie algebras" has the following definition of enveloping algebra generated by a subset (Definition 2, Chap II) : Start with an unital associative algebra $A$ (over a field $F$) and $S subset A$. The enveloping algebra $S^{ast}$ is simply the associative subalgebra of $A$ containing $1_A$ generated by $S$. The way I see it as $S^{ast} = sum_{n geq 0} S^n$ where $S^n$ is the $F$-submodule generated by the set ${ s_1 dotsc s_n ~:~ s_1, dotsc, s_n in S }$ of $n$-fold monomials (of course $S^{0} := F$). Since we require $1_A in S^{ast}$, we need $n=0$ in the sum.
Now while describing the subsequent properties, it says for $w in A$ we have ${ w }^{ast}$ is the algebra of polynomials in $w$ with constant term $0$. How can we have the constant terms $0$ if I need $1_A in { w }^{ast}$?
lie-algebras
asked Jan 20 at 17:24
SiddharthaSiddhartha
416
$endgroup$
Jacobson's book on "Lie algebras" has the following definition of enveloping algebra generated by a subset (Definition 2, Chap II) : Start with an unital associative algebra $A$ (over a field $F$) and $S subset A$. The enveloping algebra $S^{ast}$ is simply the associative subalgebra of $A$ containing $1_A$ generated by $S$. The way I see it as $S^{ast} = sum_{n geq 0} S^n$ where $S^n$ is the $F$-submodule generated by the set ${ s_1 dotsc s_n ~:~ s_1, dotsc, s_n in S }$ of $n$-fold monomials (of course $S^{0} := F$). Since we require $1_A in S^{ast}$, we need $n=0$ in the sum.
Now while describing the subsequent properties, it says for $w in A$ we have ${ w }^{ast}$ is the algebra of polynomials in $w$ with constant term $0$. How can we have the constant terms $0$ if I need $1_A in { w }^{ast}$?
lie-algebras
lie-algebras
asked Jan 20 at 17:24
SiddharthaSiddhartha
416
asked Jan 20 at 17:24
SiddharthaSiddhartha
416
edited Jan 20 at 19:30
Siddhartha
asked Jan 20 at 17:24
SiddharthaSiddhartha
416
asked Jan 20 at 17:24
SiddharthaSiddhartha
416
asked Jan 20 at 17:24
SiddharthaSiddhartha
416
416
$begingroup$
I think(?) I found the problem : while discussing the enveloping sub-algebra (containing 1) generated by $S$ the book uses the notation $S^{ast}(S^{dagger})$, elsewhere it simply uses $S^{ast}$. Maybe this is the difference between whether to take $n=0$ or not in the sum of the definition. There is no a priori clarification of these, and so I am not sure.
$endgroup$
– Siddhartha
Jan 21 at 14:12
$begingroup$
This has nothing to do with Lie algebras. It's only about associative algebras. You have added "universal" in the title but it's not called "universal" in Jacobson. Anyway, what Jacobson calls "enveloping algebra" should have been called "enveloping subalgebra", and is now rather referred to as "unital subalgebra generated by $S$".
$endgroup$
– YCor
Jan 27 at 4:13
add a comment |
$begingroup$
I think(?) I found the problem : while discussing the enveloping sub-algebra (containing 1) generated by $S$ the book uses the notation $S^{ast}(S^{dagger})$, elsewhere it simply uses $S^{ast}$. Maybe this is the difference between whether to take $n=0$ or not in the sum of the definition. There is no a priori clarification of these, and so I am not sure.
$endgroup$
– Siddhartha
Jan 21 at 14:12
$begingroup$
This has nothing to do with Lie algebras. It's only about associative algebras. You have added "universal" in the title but it's not called "universal" in Jacobson. Anyway, what Jacobson calls "enveloping algebra" should have been called "enveloping subalgebra", and is now rather referred to as "unital subalgebra generated by $S$".
$endgroup$
– YCor
Jan 27 at 4:13
$begingroup$
I think(?) I found the problem : while discussing the enveloping sub-algebra (containing 1) generated by $S$ the book uses the notation $S^{ast}(S^{dagger})$, elsewhere it simply uses $S^{ast}$. Maybe this is the difference between whether to take $n=0$ or not in the sum of the definition. There is no a priori clarification of these, and so I am not sure.
$endgroup$
– Siddhartha
Jan 21 at 14:12
$begingroup$
I think(?) I found the problem : while discussing the enveloping sub-algebra (containing 1) generated by $S$ the book uses the notation $S^{ast}(S^{dagger})$, elsewhere it simply uses $S^{ast}$. Maybe this is the difference between whether to take $n=0$ or not in the sum of the definition. There is no a priori clarification of these, and so I am not sure.
$endgroup$
– Siddhartha
Jan 21 at 14:12
$begingroup$
This has nothing to do with Lie algebras. It's only about associative algebras. You have added "universal" in the title but it's not called "universal" in Jacobson. Anyway, what Jacobson calls "enveloping algebra" should have been called "enveloping subalgebra", and is now rather referred to as "unital subalgebra generated by $S$".
$endgroup$
– YCor
Jan 27 at 4:13
$begingroup$
This has nothing to do with Lie algebras. It's only about associative algebras. You have added "universal" in the title but it's not called "universal" in Jacobson. Anyway, what Jacobson calls "enveloping algebra" should have been called "enveloping subalgebra", and is now rather referred to as "unital subalgebra generated by $S$".
$endgroup$
– YCor
Jan 27 at 4:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You've misread Jacobson (p32 in his Dover book).
In a unital associative algebra $A$, he defines $S^*$ as the non-unital subalgebra generated by $S$ and calls it "enveloping associative algebra of $S$ in $A$", and $S^dagger$ as the unital subalgebra generated by $S$ and calls it "enveloping algebra of $S$ in $A$". This is consistent with his description of $S^*$ as polynomials with 0 constant term.
(This is terrible terminology! But mathematicians from the XXth century built an irreversibly messy terminology with rings and algebras...)
answered Jan 27 at 4:23
YCorYCor
7,872929
$endgroup$
$begingroup$
I see it now : the notation $S^{ast}(S^{dagger})$ simply means these are two separate cases. It should be that way as I should have guessed, otherwise the celebrated nilpotency can never be achieved.
$endgroup$
– Siddhartha
Jan 27 at 12:36
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3080852%2fproblem-with-universal-enveloping-algebra-generated-by-a-single-element-jacobso%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You've misread Jacobson (p32 in his Dover book).
In a unital associative algebra $A$, he defines $S^*$ as the non-unital subalgebra generated by $S$ and calls it "enveloping associative algebra of $S$ in $A$", and $S^dagger$ as the unital subalgebra generated by $S$ and calls it "enveloping algebra of $S$ in $A$". This is consistent with his description of $S^*$ as polynomials with 0 constant term.
(This is terrible terminology! But mathematicians from the XXth century built an irreversibly messy terminology with rings and algebras...)
answered Jan 27 at 4:23
YCorYCor
7,872929
$endgroup$
$begingroup$
I see it now : the notation $S^{ast}(S^{dagger})$ simply means these are two separate cases. It should be that way as I should have guessed, otherwise the celebrated nilpotency can never be achieved.
$endgroup$
– Siddhartha
Jan 27 at 12:36
add a comment |
$begingroup$
You've misread Jacobson (p32 in his Dover book).
In a unital associative algebra $A$, he defines $S^*$ as the non-unital subalgebra generated by $S$ and calls it "enveloping associative algebra of $S$ in $A$", and $S^dagger$ as the unital subalgebra generated by $S$ and calls it "enveloping algebra of $S$ in $A$". This is consistent with his description of $S^*$ as polynomials with 0 constant term.
(This is terrible terminology! But mathematicians from the XXth century built an irreversibly messy terminology with rings and algebras...)
answered Jan 27 at 4:23
YCorYCor
7,872929
$endgroup$
$begingroup$
I see it now : the notation $S^{ast}(S^{dagger})$ simply means these are two separate cases. It should be that way as I should have guessed, otherwise the celebrated nilpotency can never be achieved.
$endgroup$
– Siddhartha
Jan 27 at 12:36
add a comment |
$begingroup$
You've misread Jacobson (p32 in his Dover book).
In a unital associative algebra $A$, he defines $S^*$ as the non-unital subalgebra generated by $S$ and calls it "enveloping associative algebra of $S$ in $A$", and $S^dagger$ as the unital subalgebra generated by $S$ and calls it "enveloping algebra of $S$ in $A$". This is consistent with his description of $S^*$ as polynomials with 0 constant term.
(This is terrible terminology! But mathematicians from the XXth century built an irreversibly messy terminology with rings and algebras...)
answered Jan 27 at 4:23
YCorYCor
7,872929
$endgroup$
You've misread Jacobson (p32 in his Dover book).
In a unital associative algebra $A$, he defines $S^*$ as the non-unital subalgebra generated by $S$ and calls it "enveloping associative algebra of $S$ in $A$", and $S^dagger$ as the unital subalgebra generated by $S$ and calls it "enveloping algebra of $S$ in $A$". This is consistent with his description of $S^*$ as polynomials with 0 constant term.
(This is terrible terminology! But mathematicians from the XXth century built an irreversibly messy terminology with rings and algebras...)
answered Jan 27 at 4:23
YCorYCor
7,872929
answered Jan 27 at 4:23
YCorYCor
7,872929
answered Jan 27 at 4:23
YCorYCor
7,872929
answered Jan 27 at 4:23
YCorYCor
7,872929
7,872929
$begingroup$
I see it now : the notation $S^{ast}(S^{dagger})$ simply means these are two separate cases. It should be that way as I should have guessed, otherwise the celebrated nilpotency can never be achieved.
$endgroup$
– Siddhartha
Jan 27 at 12:36
add a comment |
$begingroup$
I see it now : the notation $S^{ast}(S^{dagger})$ simply means these are two separate cases. It should be that way as I should have guessed, otherwise the celebrated nilpotency can never be achieved.
$endgroup$
– Siddhartha
Jan 27 at 12:36
$begingroup$
I see it now : the notation $S^{ast}(S^{dagger})$ simply means these are two separate cases. It should be that way as I should have guessed, otherwise the celebrated nilpotency can never be achieved.
$endgroup$
– Siddhartha
Jan 27 at 12:36
$begingroup$
I see it now : the notation $S^{ast}(S^{dagger})$ simply means these are two separate cases. It should be that way as I should have guessed, otherwise the celebrated nilpotency can never be achieved.
$endgroup$
– Siddhartha
Jan 27 at 12:36
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3080852%2fproblem-with-universal-enveloping-algebra-generated-by-a-single-element-jacobso%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
I think(?) I found the problem : while discussing the enveloping sub-algebra (containing 1) generated by $S$ the book uses the notation $S^{ast}(S^{dagger})$, elsewhere it simply uses $S^{ast}$. Maybe this is the difference between whether to take $n=0$ or not in the sum of the definition. There is no a priori clarification of these, and so I am not sure.
$endgroup$
– Siddhartha
Jan 21 at 14:12
$begingroup$
This has nothing to do with Lie algebras. It's only about associative algebras. You have added "universal" in the title but it's not called "universal" in Jacobson. Anyway, what Jacobson calls "enveloping algebra" should have been called "enveloping subalgebra", and is now rather referred to as "unital subalgebra generated by $S$".
$endgroup$
– YCor
Jan 27 at 4:13