A simple article

A. U. Thor

Opgave 7, side 82: Formuler og bevis algebraens fundamentals�tning for kvaternionske polynomier. Hvis p(h) kvaternionsk polynomium,h ∈ H, gradp(h) > 0 ⇒der findes h ∈ H :

p(h) = 0

Bevis: Lad p(x + iy + jz + kw) = u(x, y, z, w) + iv(x, y, z, w) + jr(x, y, z, w) + ks(x, y, z, w)

�nsker at vise:

(∗) = det ( \begin{matrix}u_{x} & u_{y} & u_{z} & u_{w} \\ v_{x} & v_{y} & v_{z} & v_{w} \\ r_{x} & r_{y} & r_{z} & r_{w} \\ s_{x} & s_{y} & s_{z} & s_{w} \end{matrix} ) = u_{x} ( \begin{matrix}v_{y} & v_{z} & v_{w} \\ r_{y} & r_{z} & r_{w} \\ s_{y} & s_{z} & s_{w} \end{matrix} ) − u_{y} ( \begin{matrix}v_{x} & v_{z} & v_{w} \\ r_{x} & r_{z} & r_{w} \\ s_{x} & s_{z} & s_{w} \end{matrix} ) +

u_{z} ( \begin{matrix}v_{x} & v_{y} & v_{w} \\ r_{x} & r_{y} & r_{w} \\ s_{x} & s_{y} & s_{w} \end{matrix} ) − u_{w} ( \begin{matrix}v_{x} & v_{y} & v_{z} \\ r_{x} & r_{y} & r_{z} \\ s_{x} & s_{y} & s_{z} \end{matrix} )

= u_{x}(v_{y}r_{z}s_{w} + v_{z}r_{w}s_{y} + r_{y}s_{z}v_{w} − s_{y}r_{z}v_{w} − r_{y}v_{z}s_{w} − s_{z}r_{w}v_{y})
−u_{y}(v_{x}r_{z}s_{w} + v_{z}r_{w}s_{x} + r_{x}s_{z}v_{w} − s_{x}r_{z}v_{w} − r_{x}v_{z}s_{w} − s_{z}r_{w}v_{x})
+u_{z}(v_{x}r_{y}s_{w} + v_{y}r_{w}s_{x} + r_{x}s_{y}v_{w} − s_{x}r_{y}v_{w} − r_{x}v_{y}s_{w} − s_{y}r_{w}v_{x})
−u_{w}(v_{x}r_{y}s_{z} + v_{y}r_{z}s_{x} + r_{x}s_{y}v_{z} − s_{x}r_{y}v_{z} − r_{x}v_{y}s_{z} − s_{y}r_{z}v_{x})

p′(h) = lim_{(α,β,γJ(u,v,r,s;x,y,z,w)=0⇔p′(h)=0,δ)→(0,0,0,0)} (\frac{u(x + α, y + β, z + γ, w + δ) − u(x, y, z, w)}{α + βi + γj + δk})+

i(\frac{v(x + α, y + β, z + γ, w + δ) − v(x, y, z, w)}{α + βi + γj + δk})+

j(\frac{r(x + α, y + β, z + γ, w + δ) − r(x, y, z, w)}{α + βi + γj + δk})+

k(\frac{s(x + α, y + β, z + γ, w + δ) − s(x, y, z, w)}{α + βi + γj + δk})+

\begin{matrix}(β, γ, δ) → (0, 0, 0), α → 0 & u_{x} + iv_{x} + jr_{x} + ks_{x} & 1) \\ (α, γ, δ) → (0, 0, 0), β → 0 & \frac{1}{i}u_{y} + v_{y} + \frac{j}{i} r_{y} + \frac{k}{i} s_{y} & 2) \\ (α, β, δ) → (0, 0, 0), γ → 0 & \frac{1}{j}u_{z} + \frac{i}{j} iv_{z} + r_{z} + \frac{k}{j} s_{z} & 3) \\ (α, β, γ) → (0, 0, 0), δ → 0 & \frac{1}{k} u_{w} + \frac{i}{k} iv_{w} + jr_{w} + s_{z} & 4) \end{matrix}

\begin{matrix}1) = u_{x} + iv_{x} + jr_{x} + ks_{x} \\ 2) = v_{y} − iu_{y} − js_{y} + kr_{y} \\ 3) = r_{z} + is_{z} − ju_{z} − kv_{z} \\ 4) = s_{w} − ir_{w} + jv_{w} − ku_{w} \end{matrix} ⇒ \begin{matrix}u_{x} = v_{y} = r_{z} = s_{w} \\ v_{x} = u_{y} = s_{z} = r_{w} \\ r_{x} = s_{y} = u_{z} = v_{w} \\ s_{x} = r_{y} = v_{z} = u_{w} \end{matrix}

⇒ (∗)u_{x}^{4} − u_{x}v_{x}r_{x}s_{x} + u_{x}v_{x}r_{x}s_{x} + u_{x}^{2}r_{x}^{2} + u_{x}^{2}s_{x}^{2} + u_{x}^{2}v_{x}^{2}
v_{x}^{2}u_{x}^{2} + v_{x}^{2}s_{x}^{2} + v_{x}^{2}r_{x}^{2} − v_{x}u_{x}s_{x}r_{x} + v_{x}r_{x}s_{x}u_{x} + v_{x}^{4}
−r_{x}v_{x}s_{x}u_{x} + r_{x}s_{x}u_{x}v_{x} + r_{x}^{4} + r_{x}^{2}s_{x}^{2} + r_{x}^{2}u_{x}^{2} + r_{x}^{2}v_{x}^{2}
s_{x}^{2}vx_{x}^{2} + s_{x}^{2}u_{x}^{2} + s_{x}^{2}r_{x}^{2} + s_{x}^{4} − s_{x}r_{x}u_{x}v_{x} + s_{x}v_{x}r_{x}u_{x}

= u_{x}^{4} + 2u_{x}^{2}r_{x}^{2} + 2u_{x}^{2}s_{x}^{2} + 2u_{x}^{2}v_{x}^{2} + 2v_{x}^{2}s_{x}^{2} + 2v_{x}^{2}r_{x}^{2} + v_{x}^{4} + r_{x}^{4} + s_{x}^{4} + 2r_{x}^{2}s_{x}^{2}

Dermed haves: J = 0 ⇔ \begin{matrix}u_{x} = 0 \\ v_{x} = 0 \\ r_{x} = 0 \\ s_{x} = 0 \end{matrix} ⟺^{i kraf af1)2)3)4)} p′(h) = 0 resten af beviset kan kopieres direkte fra side81_{17}−side82^{4}

Opgave 3, side 89: Vis at sf�ren S^{4n−1} har tre vektorfelter som er overalt uafh�ngige. Betragt S^{4n−1} ⊆ H^{n} ≈ R^{4} , q = (q_{1}, ..., q_{n}) ∈ S^{4n−1}, i, j, k ∈ H

Flows: θ_{1}(t, q) = (cos(t) + i sin(t))q θ_{2}(t, q) = (cos(t) + j sin(t))q θ_{3}(t, q) = (cos(t) + k sin(t))q

ξ_{1}(q) = θ_{1⋆}(\frac{d}{dt})|_{(0,q)} =^{(□)} (− sin(t) + i cos(t))|_{(0,q)} = iq

ξ_{2}(q) = θ_{2⋆}(\frac{d}{dt})|_{(0,q)} =^{(□)} (− sin(t) + j cos(t))|_{(0,q)} = jq

ξ_{3}(q) = θ_{3⋆}(\frac{d}{dt})|_{(0,q)} =^{(□)} (− sin(t) + k cos(t))|_{(0,q)} = kq

Hermed er tre overalt uafh�ngige vektorfelter fundet, hvor (□) er fundet ved

\begin{matrix}p = 1 ⇒ (i, j, k) = i \\ p = 2 ⇒ (i, j, k) = j \\ p = 3 ⇒ (i, j, k) = k \end{matrix} p = 1, 2, 3 θ_{3⋆}(\frac{d}{dt}) = D_{θ_{p}} = ∑_{i=1}^{n} \frac{dθ_{p_{i}}}{dt} |_{t=0} \frac{d}{dx_{i}} |_{q} =

∑_{i=1}^{n}(− sin(t) + (i, j, k) cos(t)q_{i}| \frac{d}{dx_{i}} |_{q} = ∑_{i=1}^{n}(i, j, k)q_{i} \frac{d}{dx_{i}} |_{q} = (i, j, k)q