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seen Jul 23 at 15:49

Jun
17
comment f is measurable iff its coordinate functions are measurable
You can take that as the definition of $\mathcal{B}(\mathbb{R})\otimes \dots \otimes \mathcal{B}(\mathbb{R})$.
Jun
17
answered f is measurable iff its coordinate functions are measurable
Jun
17
comment f is measurable iff its coordinate functions are measurable
do you know that $\mathcal{B}(\mathbb{R}^n)=\mathcal{B}(\mathbb{R})\otimes \dots \otimes \mathcal{B}(\mathbb{R})$ ?
Jun
17
comment a question on higher direct images in a product
@adrido: yes, the point is that in this point you have a global morphism $f^*(R^ig_*\mathcal{O}(b))\to (R^1\pi_1)_*(\pi_2^*\mathcal{O}(b))$ that restrict to an isomorphism on every affine open subset. Anyway, I have changed a little my answer, and now it should be more clear.
Jun
17
revised a question on higher direct images in a product
added 421 characters in body
Jun
17
comment What is the distribution of $Y_n$ and its convergency
@Michael: you're right I understand my error. For some reason that I don't know I was convinced that $E[Y_n^2]=1$ but actually we have $E[Y_n^2]=2^n$ so that en.wikipedia.org/wiki/… is not applicable. Thank you for your patience! I will cancel my previous comments for better clarity.
Jun
16
answered a question on higher direct images in a product
Jun
16
comment How do i show that $\{m/2^n:m,n\in\mathbb{R}\}$ is dense in $\mathbb{R}$?
@KajHansen Actually that property is true for $\mathbb{Z}$ that is not dense. What you need to show is that between two arbitrary points in $\mathbb{R}$ there will always be a point of the set.
Jun
16
comment pullback is injective on picard groups?
Don't you always have the zero section $\sigma\colon X \to E$ ?
Jun
16
comment a question on higher direct images in a product
Hi! Actually in this case you have $R^1{\pi_1}_*(\pi_2^* \mathcal{O}(b))=H^1(\mathbb{P}^1,\mathcal{O}(b))\otimes \mathcal{O}$ on $\mathbb{P}^1$. For example have a look at exercise 2.B of math.stanford.edu/~vakil/0708-216/216class38.pdf
Jun
16
comment Why in this case $f$ should be entire?
Yes, as long as $\Omega$ is connected.
Jun
16
comment Let I, J ideals. Are they equal?
No, because the same ideal could have two different nonreduced Groebner bases. However if the Groebner bases aren't too big you can find quite easily the corresponding reduced Groebner bases: do you know the algorithm?
Jun
16
comment Why in this case $f$ should be entire?
@FardadPouran Actually analytic and holomorphic are the same.
Jun
16
comment Hermit reciprocity, $\mathfrak{sl}_2(\mathbb{C})$
For example it follows from the Cayley-Sylvester formula (sciencedirect.com/science/article/pii/S0195669806000916) that actually tells you the decomposition of $S^n(S^mV)$ into irreducible representation. Probably there is a more direct way but I don't know it.
May
13
comment Finding expected value of E(Y^2)
Since you know the distribution of $X$ you can simply compute $$ E(X^2)=\int_{-\infty}^{+\infty} x^2 d\mathbb{P}_X = \frac{1}{9.46-1-43}\int_{1.43}^{9.46}x^2 dx $$ where $\mathbb{P}_X$ is the distribution of $X$.
Apr
15
comment Isomorphisms between finite abelian groups and cyclic groups
@LokiClock No problem.
Apr
15
comment Isomorphisms between finite abelian groups and cyclic groups
@LokiClock: actually, if $f\colon A \to B$ is an injective map between two finite sets of the same cardinality, then $|f(A)|=|A|=|B|$ so that $f(A)=B$.
Apr
15
comment Isomorphisms between finite abelian groups and cyclic groups
Regarding the surjective homomorphism $C_5\times C_5 \to C_{25}$, recall that for finite sets of the same cardinality, surjectivity is equal to injectivity.
Apr
15
awarded  Yearling
Apr
15
comment Isomorphisms in finite abelian groups
You're welcome!