Mariano Suárez-Alvarez
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8h
comment Decomposition of a homogeneous polynomial
This has the following nice generalization: If we had a multigraded domain $A=\bigoplus_{\alpha\in\mathbb Z^n}A_\alpha$, we could let $w(a)$ be the convex hull in $\mathbb R^n$ of the set of multiudegrees of its nonzerocomponents. This "width" is also multiplicative, with the sum being the Minkowski sum of convex sets.
9h
comment Decomposition of a homogeneous polynomial
One way to say this is as follows: the width of a non-zero element $a$ of a graded domain is the difference $d_{\max}(a)-d_{\min}(a)$ of the degrees of the topmost nonzero component of $a$ and that of the bottommost (?) component. Then $w$ is multiplicative: $w(a)+w(b)=w(ab)$. The result follows formally from this, just as for one-variable polynomials and the usual degree.
11h
comment Is a space compact iff it is closed as a subspace of any other space?
The empty space, of course, is closed in every topological space which contains it!
11h
comment Is a space compact iff it is closed as a subspace of any other space?
If $X$ is a non-empty topological space with topology $\tau$, then $\tau'=\{Y\}\cup\tau$ is a topology on $Y=\{X\}\cup X$ which induces on $X$ its original topology and for which $X$ is not a closed subset.
13h
comment Show that every Jordan matrix has a cyclic vector
In any case, the argument you wrote is not only not very mathematical: it does not make much sense, as far as I can tell :-|
13h
comment Show that every Jordan matrix has a cyclic vector
What exactly do you mean by a Jordan matrix?
23h
comment What's logical symbol for “for some”?
When one is writing one avoids repeating phrases all the time.
23h
comment Orbits of the symmetry group and the alternating group
You deleted a question and then reasked an essentially equal one. Please do not do that.
1d
comment Continuity Must Hold in an Entire Open Set?
You asked two questions: «is this claim correct?» and (essentially) «is this claim false?», which have opposite answers! Now people write answers which start with «Yes., blah» and it is all unnecessarily confusing
1d
comment Why does Lie algebra of affine group have two components?
Your question does not make a lot of sense, really. It has pairs simply because we use pairs to denote its elements, out of convenience, and as a consequence of the choice we made to denote the elements of the grooup as pairs. If you identify the affine group as a subgroup of $GL(n+1,\mathbb R)$ then the natural way to describe the elements of its Lie algebra would be as $(n+1)\times(n+1)$ matrices.
1d
comment A question about matrix algebras
Why is it false?
1d
comment Computation in Wikipedia's article “Riemann Curvature Tensor”
This is done in Warner's textbook in detail, if I recall correctly. Any good textbook on the subject should cover it.
2d
comment index 2 subgroups of the infinite product of Z/2Z
Every subgroup of index $2$ is a prime ideal —maximal, in fact. What you are asking, therefore, is for a descrition of the elements sof the spectrum, and you know the standard description.
2d
comment Non-finitely generated, non-divisible, non-projective, flat module, over a polynomial ring
The direct sum of two copies of my M is not a localization of R. :-)
2d
comment Non-finitely generated, non-divisible, non-projective, flat module, over a polynomial ring
There is no need to localize all variables, in fact-.
2d
answered Non-finitely generated, non-divisible, non-projective, flat module, over a polynomial ring
2d
comment Non-finitely generated, non-divisible, non-projective, flat module, over a polynomial ring
The ring of Laurent polynomials $k[x^{\pm1}]$ viewed as a$k[x]$-module satisfies your conditions in (1), no?
2d
comment Wedge product of maps: functorial vs. exterior algebra
That is not multilinear: it is not even well defined! $v_1\wedge v_2$ is the same thing as $(-v_2)\wedge v_1$, yet there is no reason for $f(v_1)\wedge g(v_2)$ to be the same thing as $f(-v_2)\wedge g(v_1)$.
2d
comment Wedge product of maps: functorial vs. exterior algebra
Your bullet point $2$ does not make sense. What you can do is to take a map $f:V\to W$ and induce from it a map $f^{\wedge k}:\Lambda^kV\to\Lambda^kW$. But if $f$ and $g$ are two maps $V\to W$, there is no such thing as $f\wedge g:V\wedge V\to W\wedge W$.
2d
comment Relation between torsion in torsion free of covariant derivative and torsion free group
No, there is no connection. ${}{}{}$