12,453 reputation
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bio website math.jussieu.fr/~leila/…
location Lhasa, Tibet
age 18
visits member for 1 year, 10 months
seen Mar 19 '13 at 2:45

We choose to do mathematics, not because it is easy, but because it is hard.

$$ \text{Einstein's Field Equations:} \quad \mathbf{G} = \frac{8 \pi G}{c^{4}} \mathbf{T}. $$


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awarded  Nice Answer
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Mar
19
comment Norms involving positive operators
A simple counterexample comes from $ -2I \leq I $. Of course, if both $ A $ and $ B $ are required to be positive, then a counterexample is a little harder to find.
Mar
19
comment Norms involving positive operators
@julien: I think saying that $ A \leq B $ requires both $ A $ and $ B $ to be at least self-adjoint operators.
Mar
19
comment Help with proving: If $X$ is a Hilbert $A$-$B$-module, then $ \| _A \langle x,x \rangle \| = \| \langle x,x \rangle _B \| $ for all $x\in X $.
@Euthenia: Do you understand the explanation given here? I hope that I’ve managed to clear your doubts. Wegge-Olsen’s book K-Theory and C$ ^{\ast} $-Algebras also contains a proof of this result.
Mar
18
revised Help finding a norm and using the Riesz Representation Theorem.
added 325 characters in body; edited title
Mar
18
revised If $ \eta $ and $ \varphi $ are closed differential forms, then prove that $ \varphi \wedge \eta $ is a closed differential form.
added 93 characters in body; edited tags; edited title
Mar
18
revised Help with proving: If $X$ is a Hilbert $A$-$B$-module, then $ \| _A \langle x,x \rangle \| = \| \langle x,x \rangle _B \| $ for all $x\in X $.
edited tags
Mar
18
comment In a C*-algebra, put $a^*a \sim aa^*$. Transitivity fails?
I get your point now, Mike. Yes, it’s important to show that $ a_{1} \in A $ because it’ll be used to establish an equivalence between $ x^{1/2} $ and $ y^{1/2} $ within $ A $ itself, and you certainly don’t want to step outside of $ A $ into $ B(\mathcal{H}) \setminus A $ in order to do that. It’s a clever argument! :) Once again, impressive job!
Mar
18
revised Extension of a continuous map on $ {\mathbf{GL}_{n}}(\mathbb{R}) $ to $ {\mathbf{M}_{n}}(\mathbb{R}) $.
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