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Robin Chapman's user avatar
Robin Chapman's user avatar
Robin Chapman's user avatar
Robin Chapman
  • Member for 14 years, 10 months
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free Z-modules: Bases etc.
I first learnt the basics of this from the textbook amazon.co.uk/Rings-Modules-Linear-Algebra-Mathematics/dp/… which now alas is out of print.
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free Z-modules: Bases etc.
I think you meant to say that the complement to the image exists iff the SNF has only zeroes and ones.
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What does it mean for a mathematical statement to be true?
Philosophers argue ad infinitum about correspondence versus coherence theories of truth.
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Computing a set of coset representatives for $\mathbb{Z}^n / \Lambda$
While (as Keith suggests) you can use the Smith normal form, you can also use the Hermite normal form. Find (using integer row operations) a generator matrix for $\Lambda$ which is upper triangular. If the diagonal entries are $d_1,\dots,d_n$ then coset reps are the $\sum a_i e_i$ where $0\le a_i < | d_i|$.
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Can Cantor set be the zero set of a continuous function?
If you have smooth compactly supported functions $f_i$ on a locally finite collection of open sets then $\sum a_i f_i$ converges to a smooth function for any $a_i$. :-) To get nice locally finite covers we need to exploit the circle of ideas around paracompactness/partitions of unity.
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Can Cantor set be the zero set of a continuous function?
Of course, a closed set need not have an upper bouned Also what is $\beta$?
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"Natural" generating sets for symmetric groups
For example (1 4) and (1 2 3 4 5 6) maybe ?
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Can Cantor set be the zero set of a continuous function?
Petya, I'm sure you're correct, but it's a bit more fiddly to ensure the uniform convergence of the sum of the derivatives of the bump functions without a convenient global coordinate system.
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Can Cantor set be the zero set of a continuous function?
It's true on any (paracompact) smooth manifold - you certainly need partitions of unity for that.
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Calculating norms over a finite field (orthogonal groups).
Isotropy of $u$ and $w$ means that $q(u)=q(w)=0$ where $q$ is the quadratic form in question. In your posting you actually discuss the case where both are nonzero. For the counting argument, think about how many pairs $(x,y)$ of elements in $\mathbb{F}_q$ satsify $xy=0$? $xy=1$? etc.
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