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Jul
5
comment Topological/homotopical classification for 1-dim CW-complexes?
Yes, and in fact, the if part is tautological, so we only need to tackle with the only if part.
Jul
3
comment Why not $SL_n (\mathbb R)$ in this exercise
There's another way to show that $A\in SL_n(\mathbb Z)\implies A^{-1}\in SL_n(\mathbb Z)$. Note that $A^{-1}=(\det A)^{-1}\operatorname{adj}(A)$ where $\operatorname{adj}(A)$ is the adjugate matrix.
Jul
3
comment Topological/homotopical classification for 1-dim CW-complexes?
Yes, but when they are removed, I cannot see any reason that two non-isomorphic graphs are not homeomorphic.
Jul
3
comment $f:\mathbb{R}\to \mathbb{R}$ continuous and $\lim_{h \to 0^{+}} \frac{f(x+2h)-f(x+h)}{h}=0$ $\implies f=$ constant.
There's only a minor gap: It only shows that $\lim_{h\to0^+}(f(x+h)-f(x))/h=0$, not $f'(x)=0$. But in this weaker condition $f$ is also a constant (It's known that if the Dini derivative $D^+f\ge0$ everywhere, then $f$ is nondecreasing).
Jul
3
comment Topological/homotopical classification for 1-dim CW-complexes?
But then are non-graph-isomorphic thrown graphs non-homeomorphic? I guess so but I cannot conceive a proof.
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jul
2
comment Topological/homotopical classification for 1-dim CW-complexes?
The difference between graph-iso and homeo is that intuitively we can merge two adjacent edges and delete the joint vertex if it's only connected with these edges. It's inequivalent, but to my intuition, it could be shown that homeo problem is somewhat no easier than graph-iso problem.
Jul
2
comment Topological/homotopical classification for 1-dim CW-complexes?
Thanks. I do have Hatcher but I didn't read it. Any idea on topological classification?
Jul
2
comment Topological/homotopical classification for 1-dim CW-complexes?
I doubt the homeo. cls. is related to graph isomorphism problem somewhat, so maybe undecidable.
Jul
2
asked Topological/homotopical classification for 1-dim CW-complexes?
Jun
27
comment If $Ax = b$ has more than one solution so does $Ax = 0$, where $A$ is $m\times n$ real matrix.
The point is that, if $Ax=0$ has a solution, since the solution set is a linear subspace, it should contain the line spanned by that solution and so infinite if the ground field is infinite.
Jun
26
accepted Simplicity is invariant under extension of scalars
Jun
25
comment Simplicity is invariant under extension of scalars
@rschwieb Prop 2,(ii) is just a generalization of the problem. Please post an answer and I'll accept it. Thanks!
Jun
25
comment Simplicity is invariant under extension of scalars
@rschwieb Where did you see the statement that if $A$ is central simple then $A\otimes_k K$ is simple? It's stated in Shafarevich's Basic notions of algebra, but claimed that it's easy to prove. That book omits many proofs (it's not a textbook but an introduction). I need some reference.
Jun
25
revised Simplicity is invariant under extension of scalars
added 68 characters in body
Jun
25
comment Simplicity is invariant under extension of scalars
@rschwieb Suppose $\sum_k a_k\otimes\xi_k\in Z(A\otimes_k K)$ and $\xi_1,\dotsc,\xi_n$ are linearly independent over $k$, and it commutes with any $b\otimes\eta$, then by simple calculation it could be shown that $a_k$ commutes with $b$ for any $b\in A$, thus $a_k\in k$ and thus $\sum_k a_k\otimes\xi_k\in K$.
Jun
24
comment Simplicity is invariant under extension of scalars
@rschwieb Well, I need to check whether that $A$ is central $k$ implies that $A\otimes_k K$ is central over $K$. I proved this, but I'm not 100% sure. I didn't touch with extension of scalars before.
Jun
24
comment Simplicity is invariant under extension of scalars
@JeremyRickard You're right. I misunderstood the sentence.
Jun
24
asked Simplicity is invariant under extension of scalars