Let $\dim V = a$ and $\dim W = b$ so that $V$ has basis $v_1, \ldots, v_a$ and $W$ has basis $w_1, \ldots, w_b$. Then $V\oplus W$ has basis elements $v_1,\ldots , v_a,w_1,\ldots ,w_b\triangleq x_1,\ldots ,x_{a+b}$. Then $\wedge^n\left(V\oplus W\right)$ has basis $$\{x_{i_1}\wedge\cdots \wedge x_{i_n}|1\leq i_1<\cdots <i_n\leq a+b\}\triangleq \\ \{x_{j_1}\wedge \cdots \wedge x_{j_r} \wedge x_{k_1} \wedge \cdots \wedge x_{k_j}|1\leq j_1 < \cdots <j_r \leq a < k_1 < \cdots <k_s \leq a+b,r+s=n\}$$
okay so this is what i'm not sure i'm understanding, I want to map these basis vectors to the appropriate summand in $\displaystyle \bigoplus_{r+s=n}\wedge^rV\otimes_k \wedge^sW$
I'm trying to find a reference for some fiber bundle stuff and I'm having a heck of a time trying to find it in some of the books I was recommended, but it might be that I don't know exactly what I'm looking for
One thing I'm trying to find a reference for was a sentence that says "Conversely, one can show that if $S^{n-1}$ is a group, it induces the (associative) normed real division algebra on $\mathbb{R}^n$"
@AlexanderGruber make bases for V and W, form a basis of $(V\oplus W)^{\otimes n}$, notice that in $\Lambda^n(V\oplus W)$ any one of these basis vectors can be rearranged to get the V vectors first and the W vectors second...
Anyways, the other question is that trying to construct octonionic projective space and trying to look at bundles like, say $\pi: S^{8n+7}\rightarrow \mathbb{O}\mathbb{P}^n$ with fibers $S^7$, you find that it apparently can't be done except for $n=1,2$, but I would like to know a bit more about why that is. This presentation I found doesn't cite anything and just states it. I wasn't sure if it was something common knowledge in fiber bundle-y stuff or not
@AlexanderGruber exterior products gives a coordinate-free definition for determinants, they give differential forms, they categorify elementary symmetric polynomials (and are special cases of schur functors, which give the irreducible GL(V)-reps of tensor powers of V, and categorify schur polynomials)
@Alexander Suppose V,W have dim=1. Pick basis vectors v and w respectively. Consider $\Lambda^3(V\oplus W)$. We know a basis for $(V\oplus W)^{\otimes 3}$ is $$\{v\otimes v\otimes v, v\otimes v\otimes w, v\otimes w\otimes v,v\otimes w\otimes w,w\otimes v\otimes v,w\otimes v\otimes w,w\otimes w\otimes v,w\otimes w\otimes w\}.$$ This is a spanning set in $\Lambda^3(V\oplus W)$.
Notice that in the exterior product, all of these vectors are among $$\{v\wedge v\wedge v,v\wedge v\wedge w,v\wedge w\wedge w,w\wedge w\wedge w\}$$ up to a $\pm$ sign by rearranging them appropriately.
@TedShifrin If $g\in {\rm GL}(V)$ then $\chi_{\Bbb S_\lambda(V)}(g)$ is the schur polynomial $s_\lambda$ in the eigenvalues of $g$. In particular if $\Bbb S_\lambda=\Lambda^n$ then $s_\lambda=e_n$ (the elementary symmetric polynomial)
@AlexanderGruber yes, but you see how the elts of the spanning set can be arranged so that each one looks like a vector in some $\Lambda^rV\otimes \Lambda^sW$. As I mentioned to Ted, doing a nontrivial example would require numbers that require too much typing.
@TedShifrin if we maintain the same ordering on the basis both from the $V,W$ comprising $V\oplus W$ and the $V$ and $W$ in $\wedge^r V, \wedge^s W$ don't we have the inverse map?
The universal property for the localization of a commutative ring $A$ over a multiplicative subset $S$ states that if $g:A\rightarrow B$ is a ring morphism such that $g(S)\subseteq B^*$, then there exists a unique ring morphism $h:S^{-1}A\rightarrow B$ with $g=h\circ i$
where $i$ is the canonical morphism
so far so good
Now, suppose I pick a multiplicative subset $T$ such that $S\subseteq T$
now if I have a morphism $g':A\rightarrow B$ such that $g'(T)\subseteq B^*$, in particular $g'(S)\subseteq B^*$, so there's a unique ring morphism from $S^{-1}A$ commuting the diagram
so $S^{-1}A\simeq T^{-1}A$ which is obviously wrong
@FernandoMartin there should be cases where the universal property for $T$ fails because $T$ isn't a subset of $C^\star$, but the universal property for $S$ succeeds because $S$ is a subset of $C^\star$. if they don't always fail and succeed in the same places they shouldn't be the same property, no?
hmm, how bout this? $f(x) = x - 1 - \log(x)$ then setting $\mathcal{D}_xf = 0 = 1 - 1/x$ implies there's an extremum at $x=1$. $\mathcal{D}_x$ of that is $1/x^2$, which is positive, so it's a local minimum. @Sush
I am curious to know this. It must not have been easy. Did he use magic for this purpose?
And why did he anagram it? Just keeping a new name would have been enough. Why did he need to maintain a link between the two names?
Is such type of magic possible?
What can be the limits to such things d...
@AwalGarg well, first you have to define what you mean by "more random". Usually, that means less of a non-random characteristic like clumping or failing some other test of randomness. If the first random number has a non-random characteristic, then the concatenation would have that same characteristic.
@AwalGarg oh, if you're hashing then the output is as random as either part, so since the passwords are not really random, the randomness comes from the salt, which is somewhat random, and definitely better than the password itself. So salting a password hash is good.
@user3123545 a good hash function produces essentially random results from different arguments, even if they are close. So no matter where the randomness is in your salted password, the salted and hashed password picks up on the randomness of the salt when hashed.
@user3123545 oh. it looked as if you were asking about the current question that AwalGarg was asking. You should link your comment to the comment you are talking about.
@user3123545 have you seen the rule $x^ax^b=x^{a+b}$? $x^a/x^b=x^{a-b}$?
@robjohn I was intending to ask you this question: Let $1,2,3,4,5,6,7,8,9,11,12,\ldots$ be the sequence of all positive integers which do not contain the digit zero. Write $\{a_n\}$ for this sequence. By comparing with a geometric series, show that $\sum\limits_{k=1}^n \dfrac{1}{a_k} < 90$. How would you do this?
Some months ago I was challenged to compute this absolutely awesome series $$\sum_{n=1}^{\infty} \log\left(1+\frac{1}{n}\right)\log\left(1+\frac{1}{2n}\right)\log\left(1+\frac{1}{2n+1}\right) $$
It's one of the most beautiful things I've ever seen. This morning I also proposed the following mind-blowing question
@Hawk Think of the sum in groups of powers of $10$: in the range $$\frac1{10^n}+\dots+\frac1{10^{n+1}-1}$$ there are $9\cdot10^n$ terms at most $\frac1{10^n}$ so the sum is at most $9$. However, only $\left(\frac9{10}\right)^n$ are counted because of the zeros. So the sum is at most $$9\left(1 +\frac9{10} +\left(\frac9{10}\right)^2 +\left(\frac9{10}\right)^3 +\dots\right)$$ which is $90$
@robjohn oh i see...I didn't realise that...okay...thank you for helping out...I have another question too...
@robjohn Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$.
@r9m I cannot say "No" because I didn't study the possibilty of relating the inverse tangent integral to my question, but I can say I had different things in mind when I created it.
@mirgee Thank you :-) Although I'm far away from having godlike skills, I can say that so far I've put very much work, love & passion in all things I did. With these things you can move the mountains.
I am curious to know this. It must not have been easy. Did he use magic for this purpose?
And why did he anagram it? Just keeping a new name would have been enough. Why did he need to maintain a link between the two names?
Is such type of magic possible?
What can be the limits to such things d...
