In the case k = 2, the definition says that
F(a v_1 + a' v_1', v_2) = a F(v_1, v_2) + a' F(v_1', v_2)
and
F(v_1, a v_2 + a' v_2') = a F(v_1, v_2) + a' F(v_1, v_2').
In other words, the map V_1 \to W,\;v_1 \mapsto F(v_1, v_2) is linear (for fixed v_2), and the map V_2 \to W,\;v_2 \mapsto...
The pullback of f: X → Z and g: Y → Z is their product in the slice category of objects over Z.
The product of X and Y is the pullback of X → 1 and Y → 1, where 1 is the terminal object.
Besides the definition of a split extension, I don't know of a general way to determine whether an extension splits or not, and as I'm not really an expert in this I don't really know what progress has been made towards the general extension problem. However, there is an interesting special case...
Let G be the quaternion group Q8 = {±1, ±i, ±j, ±k}, and let Z be its centre {1, -1}. Then 1 → Z → G → G/Z → 1 doesn't split; that is, there is no homomorphism s: G/Z → G such that ps = idG/Z (where p: G → G/Z is the quotient map). To see this, note that G/Z is abelian, so if such an s existed...
Semidirect products are easily classified. The problem is that many group extensions are not semidirect products; in fact, the extensions which are semidirect products are precisely the ones that split.
Or, by removing one handle at a time. If you have a handle, you can cut it and cap the two boundary components you get with disks. (That is, you replace a cylinder S1 × [0,1] with two disks D2 × {0,1}.) This will reduce the genus by 1, and increase the Euler characteristic by 2. Repeat until you...
A matrix is diagonal if it has no nonzero entries off the diagonal. A matrix is upper triangular if it has no nonzero entries below the diagonal. etc.
Clearly any 1x1 matrix satisfies these properties, since there are no entries off the diagonal, nonzero or not.
Why do you think this is true? Z2 x Z2 has 5 subgroups, but only 4 elements. Thus, your earlier proof doesn't appear correct.As for how to prove it, I think a simple counting argument is enough. If |G| = n, let N denote the number of subgroups of G. If d is a positive integer, then there are at...
Whoa, you're absolutely right, and I'm completely wrong. The proof fails at the triangle inequality part, and it can't be fixed.
Indeed, here's a counterexample, in a Hilbert space, even: In \ell^2, let the basis {ek} be defined by e1 = (1, 0, 0, ...), and ek = (1, 0, ..., 1/k, ...) (that is, a...
The series sure doesn't necessarily converge absolutely. However, you can show that it converges by showing that its sequence of partial sums is Cauchy. Specifically,
\left\lVert \sum_{k=m}^{n} \langle e_k, x \rangle \alpha_k e_k \right\rVert^2 \le M \sum_{k=m}^{n} \lvert \langle e_k, x \rangle...
Certainly if there exist a and b in K[X] such that af + bDf = 1, then there exist a and b in C[X] such that af + bDf = 1. This is equivalent to saying that none of the linear factors of f in C[X] are repeated (if there was such a linear factor, it would divide both f and Df).
It works because it works on basis vectors. Define the linear operator A by
Av = \sum_{i=1}^n \langle v, e_i \rangle e_i.
Then for any k,
Ae_k = \sum_{i=1}^n \langle e_k, e_i \rangle e_i
= \sum_{i=1}^n \delta_{ki} e_i = e_k.
It follows that A = I, so Av = v for any v.
By the way, there are many unbounded linear operators T that agree with A on the ek, since you can extend ek to an algebraic (i.e. Hamel) basis and define T to be whatever you want on the new basis elements. It's just that there's a unique continuous one, since the (algebraic) span of the ek is...