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 Yearling
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Jan
14
comment Embeddings into symmetric structures
I don't think this is exactly what you seek, but it is interesting and related: math.stackexchange.com/questions/1701/…
Jan
11
answered If $w'(\beta)<0$ and $\ell(w)+\ell(w')=\ell(ww')$, then $ww'(\beta)<0$?
Dec
6
comment Are you allowed to substitute for imaginary numbers on a real integral?
The $i$ is from $dx = i dc$, not from $f(ci) = i f(c)$.
Nov
6
comment The longest word in Weyl group and positive roots.
It occurs to me that I was mixing orders a bit: Given any reduced decomposition for $w$, we can order the positive roots sent negative by $w$ in the order in which the reflections sent them to a negative root. The longest element maps all positive roots to negative roots. The order of the < in my comment is such that there exists a reflection ordering equal to it. (Hence the conflation of these two orders in my first comment.)
Nov
6
comment The longest word in Weyl group and positive roots.
For two vectors in $\mathbb{R}^n$ relative to a fixed ordered basis $(a_1,\ldots,a_n) \leq_{lex} (b_1,\ldots,b_n)$ if $a_1 < b_1$ or if $a_1 = b_1, \ldots a_i = b_i$, but $a_{i+1} < b_{i+1}$. Precedence of comparison goes to the initial coordinate where the coordinates differ. (It is similar to alphabetical order, hence the name.) I believe what you are referring to is "height". Among simple roots $\alpha_1,\alpha_2$, $\alpha_1 < \alpha_2$ in lex order but are incomparable in the order you mention. The $<$ I was using is decomposition-based, but can be made to match lex order on roots.
Nov
5
comment The longest word in Weyl group and positive roots.
Section 5.2 of Bjorner and Brenti's "Combinatorics of Coxeter Groups". The order is described within the proof of Proposition 5.2.1 on page 137.
Jul
5
awarded  Yearling
Mar
31
comment Lemma regarding subgroup of group of automorphisms
The automorphism $\tau$ is applied to the coefficients of $P(x)$: If $P(x) = a_n x^n + \cdots + a_0$, then $\tau P(x) = \tau(a_n) x^n + \cdots + \tau(a_0)$. Since $\tau$ extends to a ring homomorphism from $A[x]$ to $A[x]$, we can apply $\tau$ to each factor in the product representation. The point is that $\tau P(x) = P(x)$ implies $\tau(a_i) = a_i$ for any coefficient $a_i$ of $P(x)$. Since this holds for all $\tau \in G$, the coefficients of $P(x)$ are in $A^G$.
Mar
26
answered 4D Geometry Book
Mar
22
comment What happens if singleton set is not closed
The trivial topology (only $\emptyset$ and $X$ are open) is an example of a topology on $X$ such that singleton sets are not closed, assuming $|X| > 1$.
Mar
6
answered Reference request: calculus of variations
Mar
3
revised How would I put these recurrence relation terms into a summation?
Inserted \left and \right on parentheses that needed them.
Mar
3
suggested approved edit on How would I put these recurrence relation terms into a summation?
Feb
23
comment Recognizing subgroups
With the last two examples, don't use the same ordered pair to check closure. Use $(x,y)$ and $(x', y')$ or some variant of that.
Feb
22
comment Mutual difference of vectors squared, does it have a name?
I suspect (but don't know as it wasn't me) the issue is with $\overrightarrow{v}^2$, which is not standard for any known vector operation. If you mean dot product, then $\overrightarrow{v} \cdot \overrightarrow{v}$ or $\Vert \overrightarrow{v} \Vert^2$ fixes this issue.
Feb
19
answered Name of a particular “sum of products” function
Feb
15
comment Cycle structures of $S_6$
Yes, that is probably the simplest approach.
Feb
12
comment General Classification of finite simple ternary groups?
Universal algebra provides natural definitions that generalize those you find for groups, rings, etc. "Simple algebra" is probably the definition you seek. Unlike groups and rings, the congruence relations (eq-relations preserving the operations) may not correspond to a sub-(ternary group), so there may not be a natural definition of "normal". But there will be one for "simple". Standard results in universal algebra also imply isomorphism theorems for your structure, substructure theorems, etc. (If you include the inversion operation into your signature, Birkhoff's HSP theorem applies.)
Jan
18
comment linear independence on polynomials
No. None of the polynomials have to be the zero polynomial. The hypothesis $p_j(2) = 0$ affects the coefficients of the polynomials, though it's not obvious how in the standard basis. (This is why I suggest using an alternative basis.)
Jan
18
comment linear independence on polynomials
One possibility is to use a change of basis: Instead of using $x^0,x^1,\ldots,x^m$ as a basis, you might use $(x-2)^0,(x-2)^1,\ldots,(x-2)^m$ as a basis.