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- CommentRowNumber1.
- CommentAuthorUrs
- CommentTimeMay 3rd 2010
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Author: Urs
Format: MarkdownItexedited [[Weil algebra]] a bit. More to come.

edited Weil algebra a bit. More to come.
- CommentRowNumber2.
- CommentAuthorUrs
- CommentTimeAug 16th 2010
- PermaLink
Author: Urs
Format: MarkdownItexadded to [[Weil algebra]] a section <a href="http://ncatlab.org/nlab/show/Weil+algebra#Properties">Properties</a> with a discussion of the free property of the Weil algebra.

added to Weil algebra a section Properties with a discussion of the free property of the Weil algebra.
- CommentRowNumber3.
- CommentAuthordomenico_fiorenza
- CommentTimeAug 16th 2010
- (edited Aug 16th 2010)
- PermaLink
Author: domenico_fiorenza
Format: MarkdownItexI'm not sure I'm following your argument: it seems you're saying that a dgca morphism $f:W(\mathfrak{g}) \to A$ is the same thing as a linear map $\mathfrak{g}^*\to U(A)$ which moreover satisfies the compatibility with differentials resticted to $\mathfrak{g}^*$, i.e. that argument seems to prove that $Hom_{dgca}(W(\mathfrak{g}),A)\to Hom_{grVect}(\mathfrak{g},U(A))$ is injective, not bijective. Where am I lost?

I’m not sure I’m following your argument: it seems you’re saying that a dgca morphism $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>f</mi><mo>:</mo><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo><mo>→</mo><mi>A</mi></mrow><annotation encoding="application/x-tex">f:W(\mathfrak{g}) \to A</annotation></semantics></math>$ is the same thing as a linear map $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msup><mi>𝔤</mi> <mo>*</mo></msup><mo>→</mo><mi>U</mi><mo stretchy="false">(</mo><mi>A</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">\mathfrak{g}^*\to U(A)</annotation></semantics></math>$ which moreover satisfies the compatibility with differentials resticted to $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msup><mi>𝔤</mi> <mo>*</mo></msup></mrow><annotation encoding="application/x-tex">\mathfrak{g}^*</annotation></semantics></math>$ , i.e. that argument seems to prove that $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>Hom</mi> <mi>dgca</mi></msub><mo stretchy="false">(</mo><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo><mo>,</mo><mi>A</mi><mo stretchy="false">)</mo><mo>→</mo><msub><mi>Hom</mi> <mi>grVect</mi></msub><mo stretchy="false">(</mo><mi>𝔤</mi><mo>,</mo><mi>U</mi><mo stretchy="false">(</mo><mi>A</mi><mo stretchy="false">)</mo><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">Hom_{dgca}(W(\mathfrak{g}),A)\to Hom_{grVect}(\mathfrak{g},U(A))</annotation></semantics></math>$ is injective, not bijective. Where am I lost?
- CommentRowNumber4.
- CommentAuthorUrs
- CommentTimeAug 16th 2010
- (edited Aug 16th 2010)
- PermaLink
Author: Urs
Format: MarkdownItexThe argument I gave is supposed to show that once you fix the underlying $\mathfrak{g}^* \to U(A)$, there is a unique morphism $W(\mathfrak{g}) \to A$ extending that. See, fix any images $f : \mathfrak{g}^* \to A$ of the unshifted generators $t \in \mathfrak{g}^*$. Then you can solve uniquely for the images of the shifted generators $\sigma t$ by using that in order to have a dg-homomorphism we need $$ d_A f (t) = f (d_{W} t ) = f (d_{CE} t) + f (\sigma t) $$ so that we find that $f$ has to act on the shifted generators as $$ f : \sigma t \mapsto d_A (f(t)) - f(d_{CE} t) \,. $$ So then the only remaining question is if this unique way to solve $[d,f] = 0$ on the unshifted generators also solves it on the shifted generators. This I typed out the computation for. Please let me know if this helps.

The argument I gave is supposed to show that once you fix the underlying $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msup><mi>𝔤</mi> <mo>*</mo></msup><mo>→</mo><mi>U</mi><mo stretchy="false">(</mo><mi>A</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">\mathfrak{g}^* \to U(A)</annotation></semantics></math>$ , there is a unique morphism $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo><mo>→</mo><mi>A</mi></mrow><annotation encoding="application/x-tex">W(\mathfrak{g}) \to A</annotation></semantics></math>$ extending that.

See, fix any images $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>f</mi><mo>:</mo><msup><mi>𝔤</mi> <mo>*</mo></msup><mo>→</mo><mi>A</mi></mrow><annotation encoding="application/x-tex">f : \mathfrak{g}^* \to A</annotation></semantics></math>$ of the unshifted generators $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>t</mi><mo>∈</mo><msup><mi>𝔤</mi> <mo>*</mo></msup></mrow><annotation encoding="application/x-tex">t \in \mathfrak{g}^*</annotation></semantics></math>$ .

Then you can solve uniquely for the images of the shifted generators $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>σ</mi><mi>t</mi></mrow><annotation encoding="application/x-tex">\sigma t</annotation></semantics></math>$ by using that in order to have a dg-homomorphism we need
$<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msub><mi>d</mi> <mi>A</mi></msub><mi>f</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mi>f</mi><mo stretchy="false">(</mo><msub><mi>d</mi> <mi>W</mi></msub><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mi>f</mi><mo stretchy="false">(</mo><msub><mi>d</mi> <mi>CE</mi></msub><mi>t</mi><mo stretchy="false">)</mo><mo>+</mo><mi>f</mi><mo stretchy="false">(</mo><mi>σ</mi><mi>t</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex"> d_A f (t) = f (d_{W} t ) = f (d_{CE} t) + f (\sigma t) </annotation></semantics></math>$
so that we find that $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>f</mi></mrow><annotation encoding="application/x-tex">f</annotation></semantics></math>$ has to act on the shifted generators as
$<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>f</mi><mo>:</mo><mi>σ</mi><mi>t</mi><mo>↦</mo><msub><mi>d</mi> <mi>A</mi></msub><mo stretchy="false">(</mo><mi>f</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo stretchy="false">)</mo><mo>−</mo><mi>f</mi><mo stretchy="false">(</mo><msub><mi>d</mi> <mi>CE</mi></msub><mi>t</mi><mo stretchy="false">)</mo><mspace width="0.16667em"/><mo>.</mo></mrow><annotation encoding="application/x-tex"> f : \sigma t \mapsto d_A (f(t)) - f(d_{CE} t) \,. </annotation></semantics></math>$
So then the only remaining question is if this unique way to solve $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo stretchy="false">[</mo><mi>d</mi><mo>,</mo><mi>f</mi><mo stretchy="false">]</mo><mo>=</mo><mn>0</mn></mrow><annotation encoding="application/x-tex">[d,f] = 0</annotation></semantics></math>$ on the unshifted generators also solves it on the shifted generators. This I typed out the computation for.

Please let me know if this helps.
- CommentRowNumber5.
- CommentAuthordomenico_fiorenza
- CommentTimeAug 16th 2010
- (edited Aug 16th 2010)
- PermaLink
Author: domenico_fiorenza
Format: MarkdownItexwhat I don't see is how you can have $d_A f(t)=f(d_W t)$ without imposing this condition on $f:\mathfrak{g}^*\to U(A)$. Only thing I'm able to see is that if you have this, then the unique ectension of $f$ to shifted generators is a differential on the whole $W(\mathfrak{g})$.

what I don’t see is how you can have $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>d</mi> <mi>A</mi></msub><mi>f</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mi>f</mi><mo stretchy="false">(</mo><msub><mi>d</mi> <mi>W</mi></msub><mi>t</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">d_A f(t)=f(d_W t)</annotation></semantics></math>$ without imposing this condition on $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>f</mi><mo>:</mo><msup><mi>𝔤</mi> <mo>*</mo></msup><mo>→</mo><mi>U</mi><mo stretchy="false">(</mo><mi>A</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">f:\mathfrak{g}^*\to U(A)</annotation></semantics></math>$ . Only thing I’m able to see is that if you have this, then the unique ectension of $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>f</mi></mrow><annotation encoding="application/x-tex">f</annotation></semantics></math>$ to shifted generators is a differential on the whole $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">W(\mathfrak{g})</annotation></semantics></math>$ .
- CommentRowNumber6.
- CommentAuthorUrs
- CommentTimeAug 16th 2010
- PermaLink
Author: Urs
Format: MarkdownItexWe must be misunderstanding each other on some simple point. > what I don't see is how you can have $d_A f(t)=f(d_W t)$ without imposing this condition on $f:\mathfrak{g}^*\to U(A)$. Which condition do you mean? For me $f:\mathfrak{g}^*\to U(A)$ means the images of the unshifted generators. Fix any such, without any condition. Then the equation $d_A f(t) = f(d_W t)$ has a unique solution encoded by a map $f : \mathfrak{g}^*[1] \to U(A)$: these shifted generators must be mapped as $$ f : \sigma t \mapsto d_A (f(t)) - f(d_{CE} t) \,. $$ You see, the Weil algebra is precisely built such that it introduces precisely one new generator for each condition in the Chevalley-Eilenberg algebra, precisely such that all these conditions get unique soliutions: the new generators pick up precisely the failure o the old generators to satisfy their original conditions.

We must be misunderstanding each other on some simple point.

what I don’t see is how you can have $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>d</mi> <mi>A</mi></msub><mi>f</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mi>f</mi><mo stretchy="false">(</mo><msub><mi>d</mi> <mi>W</mi></msub><mi>t</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">d_A f(t)=f(d_W t)</annotation></semantics></math>$ without imposing this condition on $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>f</mi><mo>:</mo><msup><mi>𝔤</mi> <mo>*</mo></msup><mo>→</mo><mi>U</mi><mo stretchy="false">(</mo><mi>A</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">f:\mathfrak{g}^*\to U(A)</annotation></semantics></math>$ .

Which condition do you mean? For me $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>f</mi><mo>:</mo><msup><mi>𝔤</mi> <mo>*</mo></msup><mo>→</mo><mi>U</mi><mo stretchy="false">(</mo><mi>A</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">f:\mathfrak{g}^*\to U(A)</annotation></semantics></math>$ means the images of the unshifted generators. Fix any such, without any condition. Then the equation $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>d</mi> <mi>A</mi></msub><mi>f</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mi>f</mi><mo stretchy="false">(</mo><msub><mi>d</mi> <mi>W</mi></msub><mi>t</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">d_A f(t) = f(d_W t)</annotation></semantics></math>$ has a unique solution encoded by a map $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>f</mi><mo>:</mo><msup><mi>𝔤</mi> <mo>*</mo></msup><mo stretchy="false">[</mo><mn>1</mn><mo stretchy="false">]</mo><mo>→</mo><mi>U</mi><mo stretchy="false">(</mo><mi>A</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">f : \mathfrak{g}^*[1] \to U(A)</annotation></semantics></math>$ : these shifted generators must be mapped as
$<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>f</mi><mo>:</mo><mi>σ</mi><mi>t</mi><mo>↦</mo><msub><mi>d</mi> <mi>A</mi></msub><mo stretchy="false">(</mo><mi>f</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo stretchy="false">)</mo><mo>−</mo><mi>f</mi><mo stretchy="false">(</mo><msub><mi>d</mi> <mi>CE</mi></msub><mi>t</mi><mo stretchy="false">)</mo><mspace width="0.16667em"/><mo>.</mo></mrow><annotation encoding="application/x-tex"> f : \sigma t \mapsto d_A (f(t)) - f(d_{CE} t) \,. </annotation></semantics></math>$
You see, the Weil algebra is precisely built such that it introduces precisely one new generator for each condition in the Chevalley-Eilenberg algebra, precisely such that all these conditions get unique soliutions: the new generators pick up precisely the failure o the old generators to satisfy their original conditions.
- CommentRowNumber7.
- CommentAuthordomenico_fiorenza
- CommentTimeAug 16th 2010
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Author: domenico_fiorenza
Format: MarkdownItexAh, ok! now that's perfectly clear! do you mind if I rewrite the proof on the nLab here and there? ;)

Ah, ok! now that’s perfectly clear! do you mind if I rewrite the proof on the nLab here and there? ;)
- CommentRowNumber8.
- CommentAuthorUrs
- CommentTimeAug 16th 2010
- PermaLink
Author: Urs
Format: Markdown> do you mind if I rewrite the proof on the nLab here and there? ;) Please do.

do you mind if I rewrite the proof on the nLab here and there? ;)

Please do.
- CommentRowNumber9.
- CommentAuthordomenico_fiorenza
- CommentTimeAug 16th 2010
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Author: domenico_fiorenza
Format: MarkdownItexDone. Please have a look. I've left also the old proof for the moment. One day we should take care of fixing the grading..

Done. Please have a look. I’ve left also the old proof for the moment. One day we should take care of fixing the grading..
- CommentRowNumber10.
- CommentAuthorUrs
- CommentTimeAug 16th 2010
- PermaLink
Author: Urs
Format: MarkdownThanks, looks good. I removed the redundant old paragraph now. > One day we should take care of fixing the grading. There are some comments on that at [[Chevalley-Eilenberg algebra]], at least.

Thanks, looks good. I removed the redundant old paragraph now.

One day we should take care of fixing the grading.

There are some comments on that at Chevalley-Eilenberg algebra, at least.
- CommentRowNumber11.
- CommentAuthordomenico_fiorenza
- CommentTimeAug 17th 2010
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Author: domenico_fiorenza
Format: MarkdownItexMinor edits. Added a line to explain in which sense $\sigma$ is extended to a derivation.

Minor edits. Added a line to explain in which sense $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>σ</mi></mrow><annotation encoding="application/x-tex">\sigma</annotation></semantics></math>$ is extended to a derivation.
- CommentRowNumber12.
- CommentAuthordomenico_fiorenza
- CommentTimeAug 17th 2010
- PermaLink
Author: domenico_fiorenza
Format: MarkdownItexAdded a one line construction of the canonical morphism $W(\mathfrak{g})\to CE(\mathfrak{g})$ :)

Added a one line construction of the canonical morphism $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo><mo>→</mo><mi>CE</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">W(\mathfrak{g})\to CE(\mathfrak{g})</annotation></semantics></math>$ :)
- CommentRowNumber13.
- CommentAuthordomenico_fiorenza
- CommentTimeAug 17th 2010
- (edited Aug 17th 2010)
- PermaLink
Author: domenico_fiorenza
Format: MarkdownItexHow do we see that the cohomology of $(W(\mathfrak{g}),d_{W(\mathfrak{g})})$ is trivial?

How do we see that the cohomology of $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo stretchy="false">(</mo><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo><mo>,</mo><msub><mi>d</mi> <mrow><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow></msub><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">(W(\mathfrak{g}),d_{W(\mathfrak{g})})</annotation></semantics></math>$ is trivial?
- CommentRowNumber14.
- CommentAuthorTodd_Trimble
- CommentTimeAug 17th 2010
- PermaLink
Author: Todd_Trimble
Format: MarkdownItexI am probably off base here since I am only glancing at this discussion from afar, but the construction of $W(\mathfrak{g})$ does remind me of taking the exponential (free commutative monoid) of the mapping cone of an identity morphism, which has trivial cohomology on general grounds.

I am probably off base here since I am only glancing at this discussion from afar, but the construction of $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">W(\mathfrak{g})</annotation></semantics></math>$ does remind me of taking the exponential (free commutative monoid) of the mapping cone of an identity morphism, which has trivial cohomology on general grounds.
- CommentRowNumber15.
- CommentAuthordomenico_fiorenza
- CommentTimeAug 17th 2010
- PermaLink
Author: domenico_fiorenza
Format: MarkdownItexIndeed that is what it should be! but I'm unable to switch between the general grounds definition of $W(\mathfrak{g} $ and the explicit one :(

Indeed that is what it should be! but I’m unable to switch between the general grounds definition of $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi></mrow><annotation encoding="application/x-tex">W(\mathfrak{g} </annotation></semantics></math>$ and the explicit one :(
- CommentRowNumber16.
- CommentAuthorTodd_Trimble
- CommentTimeAug 17th 2010
- PermaLink
Author: Todd_Trimble
Format: MarkdownItexOr rather, I guess the differential on $W(\mathfrak{g})$ is some twisted version of this. Anyway, what is in the back of my mind is that there is a satisfying bar construction story to be told (or fleshed out more) here which would shed light on all this.

Or rather, I guess the differential on $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">W(\mathfrak{g})</annotation></semantics></math>$ is some twisted version of this. Anyway, what is in the back of my mind is that there is a satisfying bar construction story to be told (or fleshed out more) here which would shed light on all this.
- CommentRowNumber17.
- CommentAuthorUrs
- CommentTimeAug 17th 2010
- (edited Aug 17th 2010)
- PermaLink
Author: Urs
Format: MarkdownItexYes, it comes from the mapping cone on the identity. An explicit way to see the vanishing of the cohomology is this: Let $F(\mathfrak{g})$ be the free dg-algebra on the graded vector space $\mathfrak{g}^*$, i.e. $F(\mathfrak{g}) = \wedge^\bullet (\mathfrak{g}^* \oplus \mathfrak{g}^*[1])$ with differential this shift isomorphism extended as a derivation $d_{F} = \sigma$. This manifestly has trivial cohomology. Then notice that there is a dg-algebra isomorphism $$f : F(\mathfrak{g}) \to W(\mathfrak{g}) $$ which is the identity on the unshifted genertors, and which on the shifted generators is $ f : \sigma (t) \mapsto d_W t = d_{CE} t + \sigma (t) $.

Yes, it comes from the mapping cone on the identity.

An explicit way to see the vanishing of the cohomology is this:

Let $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>F</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">F(\mathfrak{g})</annotation></semantics></math>$ be the free dg-algebra on the graded vector space $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msup><mi>𝔤</mi> <mo>*</mo></msup></mrow><annotation encoding="application/x-tex">\mathfrak{g}^*</annotation></semantics></math>$ , i.e. $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>F</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo><mo>=</mo><msup><mo>∧</mo> <mo>•</mo></msup><mo stretchy="false">(</mo><msup><mi>𝔤</mi> <mo>*</mo></msup><mo>⊕</mo><msup><mi>𝔤</mi> <mo>*</mo></msup><mo stretchy="false">[</mo><mn>1</mn><mo stretchy="false">]</mo><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">F(\mathfrak{g}) = \wedge^\bullet (\mathfrak{g}^* \oplus \mathfrak{g}^*[1])</annotation></semantics></math>$ with differential this shift isomorphism extended as a derivation $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>d</mi> <mi>F</mi></msub><mo>=</mo><mi>σ</mi></mrow><annotation encoding="application/x-tex">d_{F} = \sigma</annotation></semantics></math>$ .

This manifestly has trivial cohomology.

Then notice that there is a dg-algebra isomorphism
$<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>f</mi><mo>:</mo><mi>F</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo><mo>→</mo><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">f : F(\mathfrak{g}) \to W(\mathfrak{g}) </annotation></semantics></math>$
which is the identity on the unshifted genertors, and which on the shifted generators is

$<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>f</mi><mo>:</mo><mi>σ</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>↦</mo><msub><mi>d</mi> <mi>W</mi></msub><mi>t</mi><mo>=</mo><msub><mi>d</mi> <mi>CE</mi></msub><mi>t</mi><mo>+</mo><mi>σ</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex"> f : \sigma (t) \mapsto d_W t = d_{CE} t + \sigma (t) </annotation></semantics></math>$ .
- CommentRowNumber18.
- CommentAuthorUrs
- CommentTimeAug 17th 2010
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Author: Urs
Format: MarkdownItexI have added that remark to the Properties-section at [[Weil algebra]].

I have added that remark to the Properties-section at Weil algebra.
- CommentRowNumber19.
- CommentAuthorUrs
- CommentTimeAug 17th 2010
- PermaLink
Author: Urs
Format: MarkdownItexAlso added a remark in the section on invariant polynomials on how the dg--algebraic definition of invariant polynomial implies that these are indeed invariant under the ad-action of the $\infty$-Lie algebra.

Also added a remark in the section on invariant polynomials on how the dg–algebraic definition of invariant polynomial implies that these are indeed invariant under the ad-action of the $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mn>∞</mn></mrow><annotation encoding="application/x-tex">\infty</annotation></semantics></math>$ -Lie algebra.
- CommentRowNumber20.
- CommentAuthordomenico_fiorenza
- CommentTimeAug 17th 2010
- (edited Aug 17th 2010)
- PermaLink
Author: domenico_fiorenza
Format: MarkdownItexConcerning $F(\mathfrak{g})$ vs. $W(\mathfrak{g})$, I'd rather use the following half-line argument: given an $L_\infty$ algebra $\mathfrak{g}$, let $\mathfrak{g}_null$ be the undelying graded vector spce, which we think of as an $L_\infty$ algebra with trivial opeartions. Then the Weil algebra of $\mathfrak{g}_{null}$ is $W(\mathfrak{g}_{null})=F(\mathfrak{g})$, and the freeness property of Weil algebras tells that the identity morphism of graded vector spaces $\mathfrak{g}_{null}=\mathfrak{g}$ induces an isomorphism of differential graded commutative algebras $F(\mathfrak{g})\cong W(\mathfrak{g})$.

Concerning $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>F</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">F(\mathfrak{g})</annotation></semantics></math>$ vs. $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">W(\mathfrak{g})</annotation></semantics></math>$ , I’d rather use the following half-line argument: given an $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>L</mi> <mn>∞</mn></msub></mrow><annotation encoding="application/x-tex">L_\infty</annotation></semantics></math>$ algebra $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>𝔤</mi></mrow><annotation encoding="application/x-tex">\mathfrak{g}</annotation></semantics></math>$ , let $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>𝔤</mi> <mi>null</mi></msub></mrow><annotation encoding="application/x-tex">\mathfrak{g}_null</annotation></semantics></math>$ be the undelying graded vector spce, which we think of as an $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>L</mi> <mn>∞</mn></msub></mrow><annotation encoding="application/x-tex">L_\infty</annotation></semantics></math>$ algebra with trivial opeartions. Then the Weil algebra of $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>𝔤</mi> <mi>null</mi></msub></mrow><annotation encoding="application/x-tex">\mathfrak{g}_{null}</annotation></semantics></math>$ is $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>W</mi><mo stretchy="false">(</mo><msub><mi>𝔤</mi> <mi>null</mi></msub><mo stretchy="false">)</mo><mo>=</mo><mi>F</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">W(\mathfrak{g}_{null})=F(\mathfrak{g})</annotation></semantics></math>$ , and the freeness property of Weil algebras tells that the identity morphism of graded vector spaces $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>𝔤</mi> <mi>null</mi></msub><mo>=</mo><mi>𝔤</mi></mrow><annotation encoding="application/x-tex">\mathfrak{g}_{null}=\mathfrak{g}</annotation></semantics></math>$ induces an isomorphism of differential graded commutative algebras $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>F</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo><mo>≅</mo><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">F(\mathfrak{g})\cong W(\mathfrak{g})</annotation></semantics></math>$ .
- CommentRowNumber21.
- CommentAuthorUrs
- CommentTimeAug 17th 2010
- PermaLink
Author: Urs
Format: MarkdownItexSure, I would think this is a good extra remark to be added.

Sure, I would think this is a good extra remark to be added.
- CommentRowNumber22.
- CommentAuthordomenico_fiorenza
- CommentTimeAug 17th 2010
- PermaLink
Author: domenico_fiorenza
Format: MarkdownItexby the way, to prove that $F(\mathfrak{g})$ has trivial cohomology, I would add a line saying that $\sigma^{-1}:\mathfrak{g}^*[1]\to \mathfrak{g}^*$ extends to a degree -1 derivation $K$ of $F(\mathfrak{g})$. Since $[d_F,K]=2 Id$, multiplication by 2 is homotopic to zero, and that's all. It's completely trivial and standard, but being so short maybe it's worth adding to the nLab page.

by the way, to prove that $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>F</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">F(\mathfrak{g})</annotation></semantics></math>$ has trivial cohomology, I would add a line saying that $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msup><mi>σ</mi> <mrow><mo lspace="0.11111em" rspace="0em">−</mo><mn>1</mn></mrow></msup><mo>:</mo><msup><mi>𝔤</mi> <mo>*</mo></msup><mo stretchy="false">[</mo><mn>1</mn><mo stretchy="false">]</mo><mo>→</mo><msup><mi>𝔤</mi> <mo>*</mo></msup></mrow><annotation encoding="application/x-tex">\sigma^{-1}:\mathfrak{g}^*[1]\to \mathfrak{g}^*</annotation></semantics></math>$ extends to a degree -1 derivation $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>K</mi></mrow><annotation encoding="application/x-tex">K</annotation></semantics></math>$ of $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>F</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">F(\mathfrak{g})</annotation></semantics></math>$ . Since $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo stretchy="false">[</mo><msub><mi>d</mi> <mi>F</mi></msub><mo>,</mo><mi>K</mi><mo stretchy="false">]</mo><mo>=</mo><mn>2</mn><mi>Id</mi></mrow><annotation encoding="application/x-tex">[d_F,K]=2 Id</annotation></semantics></math>$ , multiplication by 2 is homotopic to zero, and that’s all. It’s completely trivial and standard, but being so short maybe it’s worth adding to the nLab page.
- CommentRowNumber23.
- CommentAuthorUrs
- CommentTimeAug 17th 2010
- PermaLink
Author: Urs
Format: MarkdownItexYou want me to do it?

You want me to do it?
- CommentRowNumber24.
- CommentAuthorUrs
- CommentTimeAug 17th 2010
- PermaLink
Author: Urs
Format: MarkdownItexWait a sec, it's just $[d_F, K] = Id$. (Sufficient to check this on generators.)

Wait a sec, it’s just $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo stretchy="false">[</mo><msub><mi>d</mi> <mi>F</mi></msub><mo>,</mo><mi>K</mi><mo stretchy="false">]</mo><mo>=</mo><mi>Id</mi></mrow><annotation encoding="application/x-tex">[d_F, K] = Id</annotation></semantics></math>$ .

(Sufficient to check this on generators.)
- CommentRowNumber25.
- CommentAuthordomenico_fiorenza
- CommentTimeAug 17th 2010
- PermaLink
Author: domenico_fiorenza
Format: MarkdownItexoh, yes, sure! I had just used $\sigma \sigma^{-1}+\sigma^{-1}\sigma$ without thinking that half of this expression is actually zero on generators (which half depending on which generators one considers) :) I can make the additions in a couple of hours, if you can do them before, please do. Otherwise I'll do them.

oh, yes, sure! I had just used $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>σ</mi><msup><mi>σ</mi> <mrow><mo lspace="0.11111em" rspace="0em">−</mo><mn>1</mn></mrow></msup><mo>+</mo><msup><mi>σ</mi> <mrow><mo lspace="0.11111em" rspace="0em">−</mo><mn>1</mn></mrow></msup><mi>σ</mi></mrow><annotation encoding="application/x-tex">\sigma \sigma^{-1}+\sigma^{-1}\sigma</annotation></semantics></math>$ without thinking that half of this expression is actually zero on generators (which half depending on which generators one considers) :)

I can make the additions in a couple of hours, if you can do them before, please do. Otherwise I’ll do them.
- CommentRowNumber26.
- CommentAuthorUrs
- CommentTimeAug 17th 2010
- PermaLink
Author: Urs
Format: MarkdownItex> I had just used [...] Yeah, I thought that's what you did. :-) > I can make the additions in a couple of hours, if you can do them before, please do. Okay, I did. See the last few paragraphs of the properties section. I also added a remark on what it means that $W(\mathfrak{g})$ is contractible, lest any reader is led to conclude that we are writing a long entry on just the point.

I had just used […]

Yeah, I thought that’s what you did. :-)

I can make the additions in a couple of hours, if you can do them before, please do.

Okay, I did. See the last few paragraphs of the properties section. I also added a remark on what it means that $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">W(\mathfrak{g})</annotation></semantics></math>$ is contractible, lest any reader is led to conclude that we are writing a long entry on just the point.
- CommentRowNumber27.
- CommentAuthordomenico_fiorenza
- CommentTimeAug 17th 2010
- PermaLink
Author: domenico_fiorenza
Format: MarkdownItexFine. I'm not convinced of the "elementary way" to see that Weil algebra has trivial cohomology. Namely the space $d_W(W(\mathfrak{g})$ of $d_W$-exact elements does not seems to me to be contained in $\wedge^\bullet(\mathfrak{g}^*[1])$, so a fortiori $d_W$-closed elements do not necessaily belong to that space. Maybe there's an argument which just looks at closed elements, but the "identity is homotopic to zero" seems so simple and neat to me that I would leave just that. Now I'm looking at invariant polynomials, but I have to think a while on them.

Fine. I’m not convinced of the “elementary way” to see that Weil algebra has trivial cohomology. Namely the space $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>d</mi> <mi>W</mi></msub><mo stretchy="false">(</mo><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">d_W(W(\mathfrak{g})</annotation></semantics></math>$ of $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>d</mi> <mi>W</mi></msub></mrow><annotation encoding="application/x-tex">d_W</annotation></semantics></math>$ -exact elements does not seems to me to be contained in $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msup><mo>∧</mo> <mo>•</mo></msup><mo stretchy="false">(</mo><msup><mi>𝔤</mi> <mo>*</mo></msup><mo stretchy="false">[</mo><mn>1</mn><mo stretchy="false">]</mo><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">\wedge^\bullet(\mathfrak{g}^*[1])</annotation></semantics></math>$ , so a fortiori $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>d</mi> <mi>W</mi></msub></mrow><annotation encoding="application/x-tex">d_W</annotation></semantics></math>$ -closed elements do not necessaily belong to that space. Maybe there’s an argument which just looks at closed elements, but the “identity is homotopic to zero” seems so simple and neat to me that I would leave just that.

Now I’m looking at invariant polynomials, but I have to think a while on them.
- CommentRowNumber28.
- CommentAuthorUrs
- CommentTimeAug 17th 2010
- PermaLink
Author: Urs
Format: MarkdownItexYou are right about the "elementary way". I wrote this for $F(\mathfrak{g})$, not for $W(\mathfrak{g})$, but anyway.

You are right about the “elementary way”. I wrote this for $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>F</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">F(\mathfrak{g})</annotation></semantics></math>$ , not for $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>W</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">W(\mathfrak{g})</annotation></semantics></math>$ , but anyway.
- CommentRowNumber29.
- CommentAuthordomenico_fiorenza
- CommentTimeAug 17th 2010
- (edited Aug 17th 2010)
- PermaLink
Author: domenico_fiorenza
Format: MarkdownItexyes, I meant $F(\mathfrak{g})$, but I messed up with notations :) I just wanted to point out that the differential of a product of elements in $\mathfrak{g}^*$ should not be an element of $\wedge^\bullet(\mathfrak{g}^*[1])$.

yes, I meant $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>F</mi><mo stretchy="false">(</mo><mi>𝔤</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">F(\mathfrak{g})</annotation></semantics></math>$ , but I messed up with notations :) I just wanted to point out that the differential of a product of elements in $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msup><mi>𝔤</mi> <mo>*</mo></msup></mrow><annotation encoding="application/x-tex">\mathfrak{g}^*</annotation></semantics></math>$ should not be an element of $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msup><mo>∧</mo> <mo>•</mo></msup><mo stretchy="false">(</mo><msup><mi>𝔤</mi> <mo>*</mo></msup><mo stretchy="false">[</mo><mn>1</mn><mo stretchy="false">]</mo><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">\wedge^\bullet(\mathfrak{g}^*[1])</annotation></semantics></math>$ .
- CommentRowNumber30.
- CommentAuthorUrs
- CommentTimeFeb 9th 2011
- PermaLink
Author: Urs
Format: MarkdownItexWhile on the plane to Paris, I have worked on [[Weil algebra]] a little. * Have split the definition section in two pieces, first the easier for $L_\infty$ algebras, then the genera one for $L_\infty$-algebroids; * Gave full detail in the version of $L_\infty$-algebroids of how it works over a Fermat theory and with Kähler differentials in first shifted degree. (Now at Charles-de Gaulle, about to board to Lisbon)
While on the plane to Paris, I have worked on Weil algebra a little.
- Have split the definition section in two pieces, first the easier for $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>L</mi> <mn>∞</mn></msub></mrow><annotation encoding="application/x-tex">L_\infty</annotation></semantics></math>$ algebras, then the genera one for $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>L</mi> <mn>∞</mn></msub></mrow><annotation encoding="application/x-tex">L_\infty</annotation></semantics></math>$ -algebroids;
- Gave full detail in the version of $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>L</mi> <mn>∞</mn></msub></mrow><annotation encoding="application/x-tex">L_\infty</annotation></semantics></math>$ -algebroids of how it works over a Fermat theory and with Kähler differentials in first shifted degree.
(Now at Charles-de Gaulle, about to board to Lisbon)
- CommentRowNumber31.
- CommentAuthorjim_stasheff
- CommentTimeJul 7th 2021
- PermaLink
Author: jim_stasheff
Format: TextIn re:[[CE algebra]] and [[Weil algebra]]. Add something about the restriction to finite conditions can be surpassed by using the early definitions Instead of the graded symmetric algebra on $g^*$, use Hom(the graded symmetric coalgebra on g, R) The algebroid versions use e.g. C^infty(M). What about $\Omega^*(M)\otimes g$?
In re:CE algebra and Weil algebra. Add something about the restriction to finite conditions can be surpassed by using the early definitions
Instead of the graded symmetric algebra on $g^*$, use Hom(the graded symmetric coalgebra on g, R)

The algebroid versions use e.g. C^infty(M). What about $\Omega^*(M)\otimes g$?

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