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olivier dot begassat dot cours at gmail dot com


Nov
21
comment Removing a open set from a finite open covering of a Normal space.
I don't know whether the statement is true or not.
Nov
21
comment Removing a open set from a finite open covering of a Normal space.
You will have to look for non-metrizable counter examples.
Nov
21
comment Removing a open set from a finite open covering of a Normal space.
You are asking whether an open subset of a normal space is still normal...
Nov
18
comment Another question in Bott-Tu.
Could you add the relevant details for those of us who don't have the reference handy? What is $\pi$?
Nov
17
comment Adjoint Operator of a Compact Operator
Take the time to write down an answer!
Nov
17
comment Adjoint Operator of a Compact Operator
What is the adjoint of $P_nT-T$?
Nov
17
comment Adjoint Operator of a Compact Operator
Hint: $P_n^*=P_n$.
Nov
17
comment Adjoint Operator of a Compact Operator
Are you working with Hilbert spaces? If so, you should include this hypothesis in your question.
Nov
17
answered Aut$(K/F)$ permutes roots of polynomial.
Nov
13
comment Showing $\det(I_E+f)=\sum_{p=0}^{n}\mathrm{tr}(f^p)$ where $f^p:Λ^{p}(E) \to Λ^{p}(E)$ and $f:E\to E$
You can prove this identity directly, using the matrix of $f$ relative to some basis of $E$ and explicit calculations. Or, if you are working over some real or complex vector space, you could first prove the formula on diagonalisable complex linear mappings, and conclude by density that the formula holds true for all complex linear mappings (and hence for all real linear mappings if you work over $\Bbb R$.
Nov
13
comment Showing $\det(I_E+f)=\sum_{p=0}^{n}\mathrm{tr}(f^p)$ where $f^p:Λ^{p}(E) \to Λ^{p}(E)$ and $f:E\to E$
What have you tried? BTW, this is purely linear algebra, and multilinear algebra. Differential geometry has nothing to do with this.
Nov
13
reviewed Reject suggested edit on $(n-1)$-alternative tensor on E are decomposable
Nov
13
comment How to submit a conjecture for review?
If you can back the conjecture up by a substantial amount of calculations, why not post it on MO or MSE, possibly under your real name, along with the data? If this came up while investigating a well-known conjecture, chances are your conjecture already exists, which doesn't demean it in any way, to the contrary, or has already been solved, or the claim can easily be debunked, or is a worthwhile new problem. In any case, I think everybody would benefit from this approach.
Nov
12
comment $(n-1)$-alternative tensor on E are decomposable
@JohnHughes Thanks : )
Nov
12
comment $(n-1)$-alternative tensor on E are decomposable
@mba ok, no problem. In my answer I use the standard convention, if that is confusing, I can add modifications to suit your choice of convention. What book are you using btw?
Nov
12
answered $(n-1)$-alternative tensor on E are decomposable
Nov
12
comment $(n-1)$-alternative tensor on E are decomposable
@mba Just be aware that your convention is unusual, just compare with the last sentence in this paragraph en.wikipedia.org/wiki/Exterior_algebra#Basis_and_dimension from the wikipedia article.
Nov
12
comment $(n-1)$-alternative tensor on E are decomposable
@mba There is something amiss in your question: either $\alpha\in\Lambda^{n-1}E^*$ or $\alpha\in\Lambda^{n-1}E$ and $\alpha_1,\dots,\alpha_{n-1}$ are acutally in $E$... As it stands, the question isn't correctly phrased. You are using a different convention from the standard one then, because $\Lambda^1 E\simeq E$ canonically in my book.
Nov
12
comment $(n-1)$-alternative tensor on E are decomposable
I think a star was missing, fixed it.
Nov
12
revised $(n-1)$-alternative tensor on E are decomposable
missing dual