Well, the statement $0 \ne 1$ means that $0$ is not the same thing as $1$. In other words, it means that the additive identity ($0$) is not the same as the multiplicative identity ($1$).
This statement is related to addition because it mentions the additive identity ($0$), and it's related to mu...
3 downvotes and no upvotes.
Maybe I interpreted the question too literally. I thought the asker was asking what the statement $0 \ne 1$ means. So I explained what it means.
After all, the asker wrote, "What does 0≠1 means?"
Meanwhile, the upvoted answer seems to be answering a totally different question: namely, what is the motivation behind including $0 \ne 1$ in the list of axioms?
A Serre class of abelian groups is a non-empty collection $\mathfrak{C}$ of abelian groups satisfying the property that if $0 \to A \to B \to C\to 0$ is a short exact sequence then $B \in \mathfrak{C}$ if and only if $A, C \in \mathfrak{C}$.
Note that the way you're using the term "group extension" is nonstandard, usually we say that $G$ is a group extension of $H$ (by $F$) if they fit an exact sequence $1\to F\to G\to H\to 1$
I was thinking that amenable groups have this (and others) closure property, but then I realized that all abelian groups are amenable so it's not a very interesting example
@AlessandroCodenotti The notion of Serre class is applied to homotopy groups and homology groups and most theorems about Serre classes fail when the space doesn't have abelian pi_1 acting trivially on all higher homotopy groups
@AlessandroCodenotti The key theorem is that Hurewicz theorem starts to hold modulo Serre classes. If $C$ is a Serre class, $X$ is said to be $n$-connected modulo $C$ if $\pi_i(X) \in C$ for all $0 \leq i \leq n$. Then $\pi_{n+1}(X) \cong H_{n+1}(X)$ modulo $C$, that is, the Hurewicz homomorphism $\pi_{n+1}(X) \to H_{n+1}(X)$ has kernel and cokernel in $C$
Higher homotopy and homology groups are all abelian, so that's the relevant context
Let $X$ be a Banach space and $F(X)$ the set of finite rank operators over $X$. Is $F(X)$ spanned by rank 1 projections? I think this is false, but I know it holds in Hilbert spaces
Hmmm I'm confused now. In the proof that $F(H)$ is spanned by rank one projections Murphy writes every $u\in F(H)$ as a linear combination of $n=\dim u(H)$ rank one projections
I see where my mistake was. Murphy's proof assumes wlog that $u$ is positive, so in the general case I need to put together the linear combinations I got for $u^+$ and $u^-$
While $a\otimes b$ is not positive for orthogonal $a$ and $b$
In my noob finite-dimensional brain: every matrix can be written as a nonsingular matrix times a projection. This is also known as a Hermite canonical form. Does this hold for finite-rank operators on Hilbert spaces? (Is that part and parcel of the question?)
If $H$ is a Hilbert space, $V, W \subset H$ closed subspaces, $T : V \to W$ an isomorphism between these subspaces, does there exist an extension $\widetilde{T} : H \to H$ to an isomorphism?
You just need an isomorphism of the orthocomplements
@BalarkaSen yes, looks like it. If $A$ is finite rank then let $K^\perp$ be the complement of its kernel and $P$ the projection onto that space. Let $B: K^\perp \to \mathrm{im}(A)$ be given by restriction of $A$, so $B$ is invertible. Now choose a linear iso $C$ between $K$ and $\mathrm{im}(A)^\perp$ and look at $(C\oplus B)\cdot P$
Yeah my idea was, suppose $T : H \to H$ is finite rank. Now choose a section $G : \text{im}(T) \to H$ of $T$ (easy), and extend that using the above lemma to an isomorphism $\widetilde{G} : H \to H$.
Then $TGT = T$, i.e., $G$ is a generalized inverse of $T$
$T' = GT$ is your projection
It seems for Banach you just need Hahn-Banach or something to extend instead
@AlessandroCodenotti if they have the same (Hilbert) co-dimension their complements are isomorphic $T_\perp: V^\perp\to W^\perp$, and you can define $T\oplus T_\perp$ to be a linear isomeorphism on $H$
I suppose now my question would be if this is true for compact operators as well? If $T : H \to H$ is compact, can I write $T = AP$ where $A$ is an isomorphism and $P$ is a projection?
@Semiclassical "A gap of that proof has been fixed in 2018 by V.Moretti and M.Oppio." this might be the first time I see a result by one of the professors from my bachelor in the wild
or P. A. Fillmore, On sums of projections, J. Funct. Anal. 4 (1969) 146–152. or K. Matsumoto, Self-adjoint operators as a real span of 5 projections, Math. Japon. 29 (1984) 291–294.
one says "a state on B(H) is uniquely determined by its value on projections" and the other says "every finitely additive measure on Proj(B(H)) can be extended to a state on B(H)"
well, note that the theorem in section 3 has both a forward and backwards direction
"If dim($\mathcal{H}$)≠2 then each finitely additive measure on $\mathcal{P}$ can be uniquely extended to a state on $\mathcal{B}(\mathcal{H})$. Conversely the restriction of every state to $\mathcal{P}$ is a finitely additive measure on $\mathcal{P}$."
"Theorem. Suppose $H$ is a separable Hilbert space. A measure on $H$ is a function $f$ that assigns a nonnegative real number to each closed subspace of H in such a way that, if $\textstyle \{A_{i}\}$ is a countable collection of mutually orthogonal subspaces of $H$, and the closed linear span of this collection is $B$, then $\textstyle f(B)=\sum _{i}f(A_{i})$. If the Hilbert space H has dimension at least three, then every measure f can be written in the form $\textstyle f(A)=\mathrm{Tr} (WP_{A})$, where $W$ is a positive semidefinite trace class operator and $P_{A}$ is the orthogonal proj…
@Semiclassical Let $\{e_n\}$ be a Hilbert basis and $P_{n,m}=|e_n\rangle\langle e_m|$. Then $W=\sum_{n,m} W_{n,m} P_{n,m}$ (convering weakly) and $P_{n,m}WP_{n,n}= W_{m,n}P_{n,n}$ so $Tr(WP_{n,m})=Tr(WP_{n,n}P_{n,m}) = Tr(P_{n,m}WP_{n,n}) = W_{m,n}$, thats how you can get the desnity operator
Any time you can bring two systems together in such a way that the final state of one particle depends on the input state of the other, you can make an entangled state by making that input state a quantum superposition.
Entanglement is created in a _______ manner?
a "local" manner
Is there an operator that continuously deforms a pattern into a non pattern?
@J.Doe You might ask: What are the simplest equations that will embed a shape into space? By “simplest,” mathematicians want to know two things: How many variables does it need? And what’s the highest power of those variables?
For example, the simplest equation that describes the circle requires two variables raised to the second power. Mathematicians say that its “rank” (the number of variables needed) is 2, and its “degree” (the largest exponent needed) is also 2.
so understanding the nature of the equations required to embed a shape in space also provides mathematicians with a deeper understanding of the shape itself — similar to how you could learn something about the shape of a physical object by trying to fit it inside various boxes
(@J.Doe Basically it has the advantage that since it has only "one side", both - well, one - sides equally disintegrate due to friction, increasing sustainability)
Consider $e_2, e_3, e_4 \in T_{e_1} S^3$. $\alpha \wedge d\alpha$ evaluates to $2$ on this triple, if I am not wrong. These vectors lie on the yzw-plane in $\Bbb R^4$, so out of all those terms above only $xdy \wedge dz \wedge dw$ will eat it and give something nonzero (the rest will make the parallelpiped degenerate under other coordinate plane projections). $x(e_1) = 1$, $dy \wedge dz \wedge dw(e_2, e_3, e_4) = 1$.
So I should be right, unless I am wrong, which is also likely
Hello. I asked this on math stack exchange, but it hasn't received a satisfactory answer yet. Consider the inequality lim inf x+ lim inf <= lim inf (x+y)<=lim sup(x+y)<=lim sup x + lim sup y. In each of the three inequality signs, we can replace them by either a equal sign or a strict inequality sign. That gives eight possibilities. Can each of those eight possibilities occur, and if so can anyone give me eight pairs of sequences where those possibilities occur.
The possibilities are ===, ==< , =<=, =<<, <==, <=<, <<=, <<<
@AlessandroCodenotti at the moment there is a problem, in the edit I had modified the projection $P$ (which was originally one with image $\mathcal C + \mathcal I$) to have image $\mathcal C$, now $p_i\circ P$ no longer is a projection, I'm thinking about if that can be fixed
