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seen Jul 22 at 11:36

Jul
19
comment Is a 1-form locally expressible as $dx$?
Sorry I don't understand your last comment. What I meant is that in that coordinate neighbourhood $(x_1,...,x_n)$, the dual of $\alpha$ is a vector field that can be represented as $\partial/\partial x_1$ (the rest of the vector fields that I constructed I agree won't be necessarily of the form $\partial/\partial x_i$ but I don't need that for my argument.
Jul
18
comment Is a 1-form locally expressible as $dx$?
True, but I can always find and integral curve of one of them, in this case of the one corresponding to the dual of $\alpha$. So every point will have a coordinate neighborhood in which the first coordinate gives the dual of $\alpha$. I´m not saying the same thing holds for all the others at once.
Jul
18
asked Is a 1-form locally expressible as $dx$?
Jul
17
answered Orthonormal Projection Proof
Jul
2
awarded  Curious
Feb
9
accepted Extending an independent set of vectors to a basis
Feb
6
comment Extending an independent set of vectors to a basis
I think I don't use Zorn's lemma but instead I use the axiom of choice (to construct the chain $T$) so I guess both proofs rely on the same principle (since you can get Zorn's lemma assuming the axiom of choice).
Feb
6
asked Extending an independent set of vectors to a basis
Jan
4
answered Is the finite dimension required in this proof?
Jan
4
accepted Dimension of a vector space when sum and multiplication changes
Jan
4
asked Dimension of a vector space when sum and multiplication changes
Dec
6
awarded  Nice Question
Nov
17
accepted tensor product and wedge product for direct sum decomposition
Nov
9
accepted Pfaffian system of equations
Oct
8
awarded  Tumbleweed
Oct
1
asked group of automorphism of an $SU(2)$-bundle
Sep
16
comment Are injectivity and surjectivity dual?
see also: math.stackexchange.com/questions/480837/…
Sep
16
comment Are injectivity and surjectivity dual?
Assume $w'$ is linearly independent with $w$, is neither in the image of $f$, but $w'+w$ is in the image. Then $l(w'+w)$ is not $0$ so $f^\vee(l)\neq 0$. I came with this proof: $Ann \left( Im (f) \right)= \{\varphi \in W^\vee : \varphi(w)=0 \; \forall w \in Imf \}= \{\varphi \in W^\vee : f^\vee(\varphi)=0 \} = Ker f^\vee$ So $f^\vee$ injective iff $Ker f^\vee=0$ iff $Ann \left( Im (f) \right)= 0 $ iff $Im (f)=W$ iff $f$ surjective. I'm not sure if this proof is right since as stated in Tu's book "An introduction to Manifolds" we need finite dimensions for $f$ surjective
Sep
1
comment Is the finite dimension required in this proof?
@Pink Elephants you are right!
Sep
1
comment Is the finite dimension required in this proof?
It is problem 10.5 of Tu's book "An introduction to Manifolds". Yes the axiom of choice is assumed so every vector space has a basis. @Pink Elephants the only elements which are not mapped to $0$ by $\alpha_i \circ L$ are those whose image by $L$ is a multiple of $e_i$.