Let $\sf IVect$ be the category of vector spaces with only isomorphisms. I think $V^{\otimes n}$ can be interpreted as the image of $(V,\Bbb R^n)$ under a bifunctor $\sf Vect\times IVect\to Vect$ (contravariant in the second variable). Call it $(V,W)\mapsto V^W$.
In particular, $W=\Bbb R^n$ is the standard $S_n$-rep, so $V^W$ is also an $S_n$-rep; the trivial-rep isotypic component is $S^n(V)$ and the sign-rep isotypic component is $\Lambda^n(V)$ (and more generally for arbitrary $S_n$-reps' isotypic components and Schur functors).
Hmm, wonder if $\sf Set\times Bij\to Set$ and $\sf Vect\times IVect\to Vect$ make a commutative square with the free functor $\sf Set\to Vect$ (which obviously restricts to $\sf Bij\to IVect$ as well).
@BalarkaSen There is. I wrote one down yesterday. It's only a couple of lines and does not use more than the definition of an H-space and the cohomology ring of $S^2$. The idea is just to play around what happens to the generator if one pulls it back via the multiplication map. Using the homotopy unit, one can easily derive a contradiction.
in one article the guy writes down the theorem with assertion $F$ an isometry, for a proof he gives reference of his other article and in the other article, $F$ is a periodic diffeo
why should an isometry be the same as a periodic diffeo?
$Logz=ln|z|+i Argz$ where Argz is the principal argument of z on [-π,π] the problem is that the principal argument is a multy branched function because it is the arctan(x,y) depending on the sign of x,y
principal argument is a well defined function on [-π,π] does not spit out infinite numbers also it is total differntiable as a function on $R^2$ only thing to check is the rieman cauchy eq. I can do that with brute calculations not many different cases to take.I was just wondering if it can be done fater maybe uing polar coordinates so not to examine all the cases but i dont think it is possible
An interesting discussion arose when I was talking to my former student at his REU about his project. We were talking about metrics on 3-manifolds with an $\Bbb R$-action.
@SamuelYusim More than that, I believe $V^W$ should be functorial in $V$ and $W$ (as long as the latter is restricted to the subcategory with only isomorphisms)
So you want it dependent on the vector space structure of both $V$ and $W$? And in the finite case, the only characterization of a vector space is its dimension? So $V^W$ would only be defined up to isomorphism?
I didn't mean symmetric in that way, tern. I meant that you seemed to be using only the dimension of $W$ in your example, whereas we needed to know what $V$ was as a vector space.
the reason I believe this is because the category of vector spaces with isomorphisms is equivalent to the groupoid $\bigsqcup_n GL(n)$ (viewing groups as one-object categories), and there is indeed a functor ${\sf Vect}\times\bigsqcup_n GL(n)\to \sf Vect$ satisfying these requirements
I should also mention $V^W$ should be contravariant in $W$
the reason I noticed this is because $V^{\otimes n}$ is naturally a right $GL(n)$-module. for instance when $n=2$, $$(v\otimes w)[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}]=(av+cw)\otimes(bv+dw). $$
Specifically, they want me to view the differential of $f:M\to N$ as $\mathrm df=\frac{\partial f^i}{\partial x^\alpha}\mathrm dx^\alpha\otimes \frac{\partial}{\partial f^i}$
True. I thought 0celo knew that since he asked something from Bott-Tu before where the bundle homomorphism & sections of the hom bundle thing was used.
Hello. I'm a bit confused as to these two definitions for Householder reflectors I'm getting. The first states that "A Matrix of the form: $P = I - 2uu^{T}$ is a householder reflector. And another defines it as: $F = I - 2\frac{vv^{T}}{v^{T}v}$, where $v$ is computed as: $v = sign(x_{1}) ||x||_{2} e_{1} + x$, then $v = \frac{v}{||v||_{2}}$. Are these supposed to be the same?
Because it seems that the difference is that one is divided by $v^{T}v$, and the other isn't.
Do you also see what is being done here, where they seem to use the definition of a reflector: $P = I - 2uu^{T}$ to say that an operation $Px$ is effectively equal to $x - u(2u^{T} x)$
but yeah $uu^T$ basically extracts the $u$-component of $x$, so subtracting $uu^Tx$ once from $x$ deletes the $u$-component, while subtracting it twice negates the $u$-component
maybe they put the $2$ there because they wanted to group scalars ($2$ and $u^Tx$ both being scalars)
@arctictern Do you find this strange? Universal cover of RP^2 v RP^2 is an infinite string of spheres wedged togather, not the wedge of two spheres. They are not even homotopy equivalent. Universal cover of RP^2 v S^2 is similarly wedge of three, not two, spheres. The answer then is wrong, no?
Here's another difference I've seen quite a number of times which I'm not sure what to think of. I'm using a method for calculating $u$ as: $v = sign(x_{1}) || x ||_{2} e_{1} + x$. Then $v = \frac{v}{|| v ||_{2}}$ which should be the same as $u$. However, they calculate $u$ as: $ \frac{1}{ || x - sign(x_{1}) || x ||_{2} e_{1} || } \cdot \{ \{ x_{1} - sign(x_{1}) || x ||_{2} \}, \{x_{2} \}, ... \{x_{n}\}\}$. It's not quite the same.
They subtract from vector x, I add x to my result.
If I substitute $v = sign(x_{1}) ...$ into $v = \frac{v}{|| v ||_{2}}$, Its the same except for the placement of x and the operation.
somehow I got to that notion from the fact that the alternating map $V\times\cdots\times V\to\Lambda^nV$ is ${\rm SO}(n)$invariant, where ${\rm SO}(n)$ acts on $V\times\cdots\times V\cong V\otimes {\Bbb R}^n$ from the right in the way I previously described and trivially on $\Lambda^nV$
So is: $\frac{ x - sign(x_{1}) \lVert x \rVert e_{1}}{ \lVert x - sign(x_{1}) \lVert x \rVert e_{1} \rVert }$ the same as: $\frac{ sign(x_{1}) \lVert x \rVert + x}{ \lVert sign(x_{1}) \lVert x \rVert e_{1} + x \rVert }$ ?
So imagine a mathematics teacher falls ill in the neighborhood and the school can't find a substitute, they call this firm and I have to teach these kids
Probably depends on whether you get called to work at a place called something like: Highgate Catholic Secondary School of Excellence or Trenton Juvenile Rehabilitation School for Triple Offenders.
Of course I liked teachers, but teaching isn't something I want to do. And the only teachers who tried to "impress" anything were teachers for subjects I didn't really have any interest in, like English.
Yes, but let's try to see it concretely. We have a map $f : M \to N$, $TN$ sits over $N$. Take a small chart $U$ in $N$ over which $TN$ trivializes. Then define $f^*TN$ to trivialize over $f^{-1}(U)$ (and then define the transition functions appropriately). That's what $f^*TN$ is, yeah?
You know that $U \to TN$ is $(x^1, \cdots, x^n, \partial/\partial x^1, \cdots, \partial/\partial x^n)$. Now precompose with $f : f^{-1}(U) \to U$ and rewrite the coordinates.
