Back
Here's what's floating around in my head about Quaternions and Riemann manifolds:
- Can the Quaternion basis be represented as manifold basis vectors?
- ... in $R^{1,3}$? Using +--- metric I'm assuming.
- ... in $R^3$ but with 0-forms + 1-forms?
- - what combinations of duals produces the quaternion result? What about self-duals? After reading some papers, what about anti-self-duals?
- Is quaternion basis element multiplication considered vector basis application, i.e. differentiating? If so, and the antisymmetric appear as commutation coefficients, then what about the symmetric components?
Is there even such a symmetric equivalent as commutation coefficients for Riemannian geometry? For the basis being a linear combination of the partial operator, this will always leave a second-derivative (unlike commutation which removes it).
- Is quaternion basis element multiplication the wedge product?
- Is quaternion basis element multiplication the inner product between basis elements? If so then the multiplication table is the metric tensor, and if so then we have a metric that is not symmetric nor antisymmetric, and courtesy of the $e_0 e_i = e_i e_0$ rule it isn't the case that the real are symmetric and imaginary are antisymmetric (was that the requirements for Kahler metrics? I forget.).
- - Is there an inverse metric to this metric?
Quaternions
Should I do my quaternions in a form basis? Or a vector basis? For now I'll just do vectors and use a +++ spatial basis with +1 signature.
Quaternions with a vector basis:
$q = q^0 e_0 + q^1 e_1 + q^2 e_2 + q^3 e_3$.
$q = q^0 e_0 + q^i e_i$
Quaternion multiplication table:
$e_\mu \cdot e_\nu =$
$\begin{array}{c|cccc}
& e_0 & e_1 & e_2 & e_3 \\
\hline
e_0 & e_0 & e_1 & e_2 & e_3 \\
e_1 & e_1 &-e_0 & e_3 &-e_2 \\
e_2 & e_2 &-e_3 &-e_0 & e_1 \\
e_3 & e_3 & e_2 &-e_1 &-e_0 \\
\end{array}$
One way to represent the multiplication table is to be summed up as $e_0 \cdot e_\mu = e_\mu$
and $e_i \cdot e_j = \epsilon_{ijk} e_k - \delta_{ij} e_0$
for Greek indexes ranging (0,1,2,3) and Latin indexes ranging (1,2,3).
So how can we represent this concisely?
In 4D basis: $e_\mu \cdot e_\nu = 2 \delta_{0 ( \mu} e_{\nu)} - \delta_{\mu\nu} e_0 + \epsilon_{0 \mu \nu \rho} e_\rho$
Ok of course there's the common-sense way to do this, to truly use the inner product $e_\mu \cdot e_\nu$ as the table above states.
This of course would mean that the table above is the metric.
And this of course would mean that the metric isn't symmetric.
So maybe we can get around its lack of symmetry by claiming that the $e_i$ truly are "imaginary" basii off into separate dimensions, and not just that they are negative-signature.
If this were true then we would almost be able to argue that our metric's non-real components were skew-symmetric, like a Kahler metric would be -- it is for the non-real x non-real components -- but it is not skew-symmetric among the real x non-real components.
So yeah we have a metric I guess of:
$e_\mu \cdot e_\nu = g_{\mu\nu} = \left[ \begin{matrix}
e_0 & e_1 & e_2 & e_3 \\
e_1 &-e_0 & e_3 &-e_2 \\
e_2 &-e_3 &-e_0 & e_1 \\
e_3 & e_2 &-e_1 &-e_0 \\
\end{matrix} \right]$
The metric is, however, skew-symmetric among its non-$e_0$ components. So it's halfway there. Either way I'm sure there's something about Hyper-Kahler or Quaternion-ic metrics or something.
I wanted to poke this bear with the classic notions of reals rather than just immediately jumping into the world of Cayley-Dickson constructs immediately.
Ok so inverse metric? Such that $g_{\gamma\beta} g^{\alpha\gamma} = \delta^\alpha_\beta$. Mind you I'm dealing with anti-commutative objects so better check left and right multiplication of the inverse with the metric, and maybe how conjugates work.
$1 = g_{00} g^{00} + g_{01} g^{10} + g_{02} g^{20} + g_{03} g^{30} = e_0 g^{00} + e_1 g^{10} + e_2 g^{20} + e_3 g^{30}$
$0 = g_{10} g^{00} + g_{11} g^{10} + g_{12} g^{20} + g_{13} g^{30} = e_1 g^{00} - e_0 g^{10} + e_3 g^{20} - e_2 g^{30}$
$g^{\mu\nu} = \frac{1}{4} \left[ \begin{matrix}
... TODO
\end{matrix} \right]$
$g_{\mu\nu} g^{\nu\alpha} = \delta_\mu^\alpha$
$\delta_\mu^\alpha = (2 \delta_{0 ( \mu} e_{\nu)} - \delta_{\mu\nu} e_0 + \epsilon_{0 \mu \nu \rho} e_\rho) g^{\nu\alpha}$
$\delta_\mu^\alpha =
\delta_{0 \mu} e_{\nu} g^{\nu \alpha}
+ e_\mu g^{0 \alpha}
- e_0 g^{\mu\alpha}
+ \epsilon_{0 \mu \nu \rho} e_\rho g^{\nu \alpha}
$
...split... for $\mu = 0, \alpha = 0$
$1 =
e_{\nu} g^{\nu 0}
+ e_0 g^{00}
- e_0 g^{00}
+ \epsilon_{0 0 \nu \rho} e_\rho g^{\nu 0}
$
$1 = e_\nu g^{\nu 0}$
... for $\nu = 0$:
$1 = e_0 g^{0 0}$
so $g^{00} = e_0$
... for $\nu = i$:
$1 = e_i g^{i 0}$
so $g^{i 0} = -e_i$
... for $\mu = i, \alpha = 0$
$0 =
e_i g^{0 0}
- e_0 g^{i 0}
+ \epsilon_{i j k} e_k g^{j 0}
$
$0 =
e_i e_0
+ e_0 e_i
- \epsilon_{i j k} e_k e_j
$
$0 = 3 e_i$
hmmm...
TODO am I stuck?
What if, for
$g_{\mu\nu} = e_\mu \cdot e_\nu$,
then
$g^{\mu\nu} = \frac{1}{4} \overline{e_\mu} \cdot \overline{e_\nu}$
Then $g_{\alpha\mu} g^{\mu\beta} = e_\alpha \cdot e_\mu \cdot \frac{1}{4} \overline{e_\mu} \cdot \overline{e_\beta}$
... using $\frac{1}{4} \Sigma_\mu e_\mu \cdot \overline{e_\mu} = \frac{1}{4} ( (e_0)^2 - (e_1)^2 - (e_2)^2 - (e_3)^2 ) = \frac{1}{4} (1+1+1+1) = 1$
$= e_\alpha \cdot \overline{e_\beta}$
Now this does give $\delta_{\alpha\beta}$ when $\alpha = \beta$, but it gives other non-real values otherwise.
How to get rid of them all?
If they are non-real then we can add the conjugate of the result and elminiate all the non-reals:
$\frac{1}{2} (e_\mu + \overline{e_\mu}) = 1 $ for $\mu = 0$, 0 otherwise.
So that means $\delta_{\alpha\beta} = \frac{1}{2} (
(e_\alpha \cdot e_\mu) \cdot ( \overline{e_\mu} \cdot \overline{e_\beta} )
+ \overline{ (e_\alpha \cdot e_\mu) \cdot ( \overline{e_\mu} \cdot \overline{e_\beta} ) }
)$
$ = \frac{1}{2} (
(e_\alpha \cdot e_\mu) \cdot ( \overline{e_\mu} \cdot \overline{e_\beta} )
+ \overline{ ( \overline{e_\mu} \cdot \overline{e_\beta} ) } \cdot \overline{ (e_\alpha \cdot e_\mu) }
)$
$ = \frac{1}{2} (
(e_\alpha \cdot e_\mu) \cdot ( \overline{e_\mu} \cdot \overline{e_\beta} )
+ (e_\beta \cdot e_\mu) \cdot (\overline{e_\mu} \cdot \overline{e_\alpha} )
)$
So it looks like $\frac{1}{2} (g_{\alpha\mu} g^{\mu\beta} + g_{\beta\mu} g^{\mu\alpha}) = \delta_\alpha^\beta$
What this means is that $g_{\alpha\mu} g^{\mu\beta}$ is self-adjoint.
Now with this as our metric, we don't necessarily need any commutation. So I'm going to set $[e_\mu, e_\nu] = e_\mu(e_\nu(\phi)) - e_\nu(e_\mu(\phi)) = 0$.
TODO consider quaternion multiplication as successive derivative operators.
What does quaternion multiplcation look like?
$q_1 \cdot q_2$
$= ({q_1}^0 e_0 + {q_1}^i e_i) \cdot ({q_2}^0 e_0 + {q_2}^j e_j)$
$= {q_1}^0 e_0 \cdot ({q_2}^0 e_0 + {q_2}^j e_j) + {q_1}^i e_i \cdot ({q_2}^0 e_0 + {q_2}^j e_j)$
$= {q_1}^0 e_0 \cdot {q_2}^0 e_0
+ {q_1}^0 e_0 \cdot {q_2}^j e_j
+ {q_1}^i e_i \cdot {q_2}^0 e_0
+ {q_1}^i e_i \cdot {q_2}^j e_j
$
... apply quaternion basis multiplication rules ...
$= {q_1}^0 {q_2}^0 e_0
+ {q_1}^0 {q_2}^i e_i
+ {q_1}^i {q_2}^0 e_i
+ {q_1}^i {q_2}^j (\epsilon_{ijk} e_k - \delta_{ij} e_0)
$
$= ({q_1}^0 {q_2}^0 - {q_1}^i {q_2}^j \delta_{ij}) e_0
+ (
{q_1}^0 {q_2}^k
+ {q_2}^0 {q_1}^k
+ {q_1}^i {q_2}^j \epsilon_{ijk}
) e_k
$
So how do we represent multiplication in a differential geometry framework?
We could consider quaternion-multiply to be vector-application.
Then we would be looking at the commutation coefficients, but this is only antisymmetric:
${c_{\mu\nu}}^\rho e_\rho = [e_\mu, e_\nu] = \epsilon_{0 \mu \nu \rho} e_\rho$
We could do symmetric products, but only symmetric:
$(e_\mu, e_\nu) = 2 \delta_{0 (\mu} e_{\nu)} - \delta_{\mu\nu} e_0$
In both cases the vector-application is going to produce derivatives in addition to the commutations, something I want to avoid.
Quaternions as 0-forms + 1-forms in R3
So instead, how about combining the "real" and "imaginary" components using a 0-form and a 1-form in 3-dimensions:
$q = q_0 + q_1 e^1 + q_2 e^2 + q_3 e^3$
$q = q_0 + q_i e^i$
$ \star q = \star q_0 + q_i \star e^i = q_0 \star 1 + \frac{1}{2} q_i {\epsilon^i}_{jk} e^j \wedge e^k$
Useful:
In 3-dimensions, $\epsilon^i \wedge \epsilon^j \wedge \epsilon^k = \epsilon^{ijk} \star 1$
Useful:
$e^i \wedge \star e^j$
$= e^i \wedge \frac{1}{2} {\epsilon^j}_{kl} e^k \wedge e^l$
$= \frac{1}{2} g^{jm} \epsilon_{mkl} e^i \wedge e^k \wedge e^l$
$= \frac{1}{2} g^{jm} \epsilon_{mkl} \epsilon^{ikl} \star 1$
$= g^{ij} \star 1$
Mind you $e^i$ is a 1-form and $\star e^j$ is a 2-form, so the wedge product is symmetric:
$e^i \wedge \star e^j = \star e^j \wedge e^i = g^{ij} \star 1$
Wedge product of quaterion #1 and dual of quaternion #2. This is the definition of the dot product for any form, but mind you this is a combined 0-form and 1-form so it doesn't fit with that dot-product definition:
$q^1 \wedge \star q^2$
$= ({q^1}_0 + {q^1}_i e^i) \wedge \star ({q^2}_0 + {q^2}_j e^j)$
$ = ({q^1}_0 + {q^1}_i e^i) \wedge ({q^2}_0 \star 1 + {q^2}_j \star e^j)$
$ = {q^1}_0 \wedge {q^2}_0 \star 1
+ {q^1}_0 \wedge {q^2}_j \star e^j
+ {q^1}_i e^i \wedge {q^2}_0 \star 1
+ {q^1}_i e^i \wedge {q^2}_j \star e^j
$
... using $e^i \wedge \star 1 = 0$...
$ = {q^1}_0 {q^2}_0 \star 1
+ {q^1}_0 {q^2}_j \star e^j
+ {q^1}_i {q^2}_j e^i \wedge \star e^j
$
... using $e^i \wedge \star e^j = g^{ij} \star 1$ ...
$ = {q^1}_0 {q^2}_0 \star 1
+ {q^1}_0 {q^2}_j \star e^j
+ {q^1}_i {q^2}_j g^{ij} \star 1
$
$ = ({q^1}_0 {q^2}_0 + {q^1}_i {q^2}_j g^{ij}) \star 1
+ {q^1}_0 {q^2}_j \star e^j
$
Hmm I was hoping I could isolate the inner product of the vectors, so ... yeah ... that's not looking good ...
Well what about the wedge of the dual and dual?
Wedge product of dual of #1 and dual of #2:
$\star q^1 \wedge \star q^2$
$= \star ({q^1}_0 + {q^1}_i e^i) \wedge \star ({q^2}_0 + {q^2}_j e^j)$
$= ({q^1}_0 \star 1 + {q^1}_i \star e^i) \wedge ({q^2}_0 \star 1 + {q^2}_j \star e^j)$
$=
{q^1}_0 \star 1 \wedge {q^2}_0 \star 1
+ {q^1}_0 \star 1 \wedge {q^2}_j \star e^j
+ {q^1}_i \star e^i \wedge {q^2}_0 \star 1
+ {q^1}_i \star e^i \wedge {q^2}_j \star e^j
$
$= 0$
So that won't help.
Of coure we can work $a \wedge \star b$ into an inner product with our isolating-k-forms trick of wedging, dualing, wedging to elimiate everything we don't want.
In this case if we want the $\star 1$ term then we can just dual and wedge by $\star 1$ again to eliminate that extra $\star e^j$ term:
$\star (q^1 \wedge \star q^2)$
$ = \sigma({q^1}_0 {q^2}_0 + {q^1}_i {q^2}_j g^{ij})
+ \sigma {q^1}_0 {q^2}_j e^j
$
$\star 1 \wedge \star (q^1 \wedge \star q^2)$
$ = \sigma ({q^1}_0 {q^2}_0 + {q^1}_i {q^2}_j g^{ij}) \star 1$
... and now the extra term is gone.
$\star (\star 1 \wedge \star (q^1 \wedge \star q^2))$
$ = {q^1}_0 {q^2}_0 + {q^1}_i {q^2}_j g^{ij}$
... and now we've got something close to our inner product. Thanks to the (0+1)/3 space we now have symmetric ${q^1}_0 {q^2}_i + {q^1}_i {q^2}_0$ (unlike in the 1/4 case below), buuut,
now our metric is decoupled from the time component.
Wedge product of two quaternions:
$q^1 \wedge q^2$
$= ({q^1}_0 + {q^1}_i e^i) \wedge ({q^2}_0 + {q^2}_j e^j)$
$= (({q^1}_0 + {q^1}_i e^i) \wedge {q^2}_0 + ({q^1}_0 + {q^1}_i e^i) \wedge {q^2}_j e^j)$
$= {q^1}_0 \wedge {q^2}_0 + {q^1}_i e^i \wedge {q^2}_0 + {q^1}_0 \wedge {q^2}_j e^j + {q^1}_i e^i \wedge {q^2}_j e^j$
$= {q^1}_0 {q^2}_0 + {q^2}_0 {q^1}_i e^i + {q^1}_0 {q^2}_j e^j + {q^1}_i {q^2}_j e^i \wedge e^j$
$= {q^1}_0 {q^2}_0 + ( {q^2}_0 {q^1}_i + {q^1}_0 {q^2}_i ) e^i + {q^1}_i {q^2}_j {\epsilon^{ij}}_k \star e^k$
So here we have all our components of quaternion multiplication, except our vector cross vector term is in the dual space instead of the vector space,
and we are also missing the $-{q^1}_i {q^2}_j \delta^{ij}$ real term. But some foreshadowing, we're really not interested in that first ${q^1}_0 {q^2}_0$ term. So getting rid of that...
$\star (q^1 \wedge q^2)
= {q^1}_0 {q^2}_0 \star 1 + ( {q^2}_0 {q^1}_i + {q^1}_0 {q^2}_i ) \star e^i + {q^1}_i {q^2}_j {\epsilon^{ij}}_k \sigma e^k$
$\star (q^1 \wedge q^2) \wedge e^l
= {q^1}_0 {q^2}_0 \star 1 \wedge e^l
+ ( {q^2}_0 {q^1}_i + {q^1}_0 {q^2}_i ) \star e^i \wedge e^l
+ {q^1}_i {q^2}_j {\epsilon^{ij}}_k \sigma e^k \wedge e^l
$
$= ( {q^2}_0 {q^1}^l + {q^1}_0 {q^2}^l )
+ {q^1}_i {q^2}_j {\epsilon^{ij}}_k \sigma {{\epsilon^k}_m}^l \star e^m
$
$= ( {q^2}_0 {q^1}^l + {q^1}_0 {q^2}^l )
+ {q^1}_i {q^2}_j \delta^{ij}_{mn} \sigma g^{ln} \star e^m
$
$\star (q^1 \wedge q^2) \wedge e^l \wedge e^p g_{lp}
= ( {q^2}_0 {q^1}^l + {q^1}_0 {q^2}^l ) \wedge e^p g_{lp}
+ {q^1}_i {q^2}_j \delta^{ij}_{mn} \sigma g^{ln} \star e^m \wedge e^p g_{lp}
$
$= ( {q^2}_0 {q^1}_p + {q^1}_0 {q^2}_p ) e^p
+ {q^1}_i {q^2}_j \delta^{ij}_{mp} \sigma \star e^m \wedge e^p
$
TODO does this cancel? don't do that ...
$= ( {q^2}_0 {q^1}_p + {q^1}_0 {q^2}_p ) e^p
+ {q^1}_i {q^2}_j \delta^{ij}_{mp} g^{mp}
$
TODO did I lose a dual here?
$= ( {q^2}_0 {q^1}_i + {q^1}_0 {q^2}_i ) e^i
+ {q^1}_i {q^2}_j {\epsilon^{ij}}_k e^k
$
Now if we combine this with the wedge of dual:
$q^1 \wedge q^2 + q^1 \wedge \star q^2$
$= {q^1}_0 {q^2}_0
+ ( {q^2}_0 {q^1}_i + {q^1}_0 {q^2}_i ) e^i
+ {q^1}_i {q^2}_j {\epsilon^{ij}}_k \star e^k
+ ({q^1}_0 {q^2}_0 + {q^1}_i {q^2}_j g^{ij}) \star 1
+ {q^1}_0 {q^2}_j \star e^j
$
We have some extra terms. We have that last ${q^1}_0 {q^2}_j \star e^j$. And then if you do look at self-duals, we have ${q^1}_0 {q^2}_0$ appearing twice. How to fix. By removing those isolated terms before.
$\star (\star 1 \wedge \star (q^1 \wedge \star q^2)) = {q^1}_0 {q^2}_0 + {q^1}_i {q^2}_j g^{ij}$
$\star (q^1 \wedge q^2) \wedge e^k \wedge e^l g_{kl} = ( {q^2}_0 {q^1}_i + {q^1}_0 {q^2}_i ) e^i + {q^1}_i {q^2}_j {\epsilon^{ij}}_k e^k$
So
$\star (\star 1 \wedge \star (q^1 \wedge \star q^2)) + \star (q^1 \wedge q^2) \wedge e^k \wedge e^l g_{kl}
= {q^1}_0 {q^2}_0 + {q^1}_i {q^2}_j g^{ij} + ( {q^2}_0 {q^1}_i + {q^1}_0 {q^2}_i ) e^i + {q^1}_i {q^2}_j {\epsilon^{ij}}_k e^k$
And for the case that $g^{ij} = diag(-1,-1,-1)$ we have reproduced quaternion multiplication.
FIXME use a form basis
So how would quaternion-multiplication work?:
Mind you if we self-dual'd the result then we could argue that all the components are where they are supposed to be.
$H(q) := \frac{1}{2} (q + \star q)$
$H(q_1 \wedge q_2)
= {q_1}^0 {q_2}^0 (\frac{1}{2} (1 + \star 1))
+ (
{q_2}^0 {q_1}^i
+ {q_1}^0 {q_2}^i
+ {q_1}^j {q_2}^j \epsilon_{ijk}
) (\frac{1}{2} (e_i + \star e_i))
$
we have one unique scalar and vector.
So how can we map all my 2-forms back to my 1-forms?
Or in the self-dual case, how do we simply remove the 2-forms and 3-forms?
Here's how to map 0-forms + 3-forms $(a + b \star 1)$ back to 0-forms a:
...using $\star 1 \wedge \star 1 = 0$ ...
$\star 1 \wedge (a + b \star 1)$
$= a \star 1 + b \star 1 \wedge \star 1$
$= a \star 1$
Therefore, for metric signature $\sigma$...
$\sigma \star (\star 1 \wedge (a + b \star 1))$
$= \sigma \star (a \star 1)$
$= \sigma^2 a$
$= a$
Same with mapping to the 3-form, just use the dual:
$\star 1 \wedge \star (a + b \star 1) = \star 1 \wedge (\sigma b + a \star 1) = \sigma b \star 1$
$\star (\star 1 \wedge \star (a + b \star 1)) = \sigma b \star \star 1 = \sigma^2 b = b$
So back to our "inner-product":
$\star (\star 1 \wedge \star (q_1 \wedge \star q_2))$
$ = \star (\star 1 \wedge \star (
({q_1}^0 {q_2}^0 + {q_1}^i {q_2}^j \delta_{ij}) \star 1
+ {q_1}^0 {q_2}^j \star e_j
))$
$ = \sigma \star (\star 1 \wedge (
({q_1}^0 {q_2}^0 + {q_1}^i {q_2}^j \delta_{ij})
+ {q_1}^0 {q_2}^j e_j
))$
$ = \sigma \star (\star 1
({q_1}^0 {q_2}^0 + {q_1}^i {q_2}^j \delta_{ij})
)$
$ = {q_1}^0 {q_2}^0 + {q_1}^i {q_2}^j \delta_{ij}$
But what about mapping 1-forms + 2-forms onto just 1-forms?
$w = a^i e_i + \frac{1}{2} b^i \epsilon_{ijk} e_j \wedge e_k$
so $e_l \wedge w$
$= e_l \wedge (a^i e_i + \frac{1}{2} b^i \epsilon_{ijk} e_j \wedge e_k)$
$= a^i e_l \wedge e_i + \frac{1}{2} b^i \epsilon_{ijk} e_l \wedge e_j \wedge e_k$
$= a^i \epsilon_{lim} \star e_m + \frac{1}{2} b^i \epsilon_{ijk} \epsilon_{ljk} \star 1$
$= a^i \epsilon_{lim} \star e_m + b^i \delta_{il} \star 1$
$= a^i \epsilon_{lim} \star e_m + b^l \star 1$
so $e_n \wedge e_l \wedge w$
$= e_n \wedge (
a^i \epsilon_{lim} \star e_m + b^l \star 1
)$
$= (
a^i \epsilon_{lim} e_n \wedge \star e_m + b^l e_n \wedge \star 1
)$
... $e_n \wedge \star 1 = 0$ ...
$= a^i \epsilon_{lim} e_n \wedge \star e_m$
$= \frac{1}{2} a^i \epsilon_{lim} \epsilon_{mpq} e_n \wedge e_p \wedge e_q$
$= \frac{1}{2} a^i \delta^{li}_{pq} e_n \wedge e_p \wedge e_q$
$= \frac{1}{2} a^i (\delta^l_p \delta^i_q - \delta^l_q \delta^i_p) e_n \wedge e_p \wedge e_q$
$= \frac{1}{2} a^i \delta^l_p \delta^i_q e_n \wedge e_p \wedge e_q
- \frac{1}{2} a^i \delta^l_q \delta^i_p e_n \wedge e_p \wedge e_q$
$= \frac{1}{2} a^i e_n \wedge e_l \wedge e_i
- \frac{1}{2} a^i e_n \wedge e_i \wedge e_l$
$= \frac{1}{2} a^i e_n \wedge e_l \wedge e_i
+ \frac{1}{2} a^i e_n \wedge e_l \wedge e_i$
$= a^i e_n \wedge e_l \wedge e_i$
so $\star (e_n \wedge e_l \wedge w)$
$= \star (a^i e_n \wedge e_l \wedge e_i)$
$= a^i \star (e_n \wedge e_l \wedge e_i)$
$= a^i \epsilon_{nli}$
So if we choose to do our double-wedge with $\star (\star e_j \wedge w)$ then we should get back $a^j$.
And if we want a form out of that, just do $e_j \wedge \star (\star e_j \wedge w) = a^j e_j$
Maybe I can use this trick to isolate my vector inner product and complete my quaternion multiplication?
Sum of all k-forms in 0+3 wedge product space:
$w = a + b_i e^i + c_i \star e^i + d \star 1$
Isolate the 0-form:
$\star 1 \wedge w = \star 1 a + $ ... lots of terms that turn into zero ...
$\sigma \star (\star 1 \wedge w) = \sigma \star \star 1 a = \sigma^2 a = a$
Isolate the 3-form:
$\star (\star 1 \wedge \star w) = \star (\star 1 \sigma d) = \sigma^2 d = d$
Isolate the 1-form:
$\star e^j \wedge w$
$= \star e^j \wedge (a + b_i e^i + c_i \star e^i + d \star 1)$
$= a \star e^j + b_i \star e^j \wedge e^i$
$= a \star e^j + b_j \star 1$
so $\sigma \star (\star e^j \wedge w) = a e^j + b_j$
so $\star 1 \wedge \sigma \star (\star e^j \wedge w) = b_j \star 1$
so $\sigma \star (\star 1 \wedge \sigma \star (\star e^j \wedge w)) = b_j$
so $\delta_{ij} e^i \wedge \sigma \star (\star 1 \wedge \sigma \star (\star e^j \wedge w)) = b_j e^j$
so $\delta_{ij} e^i \wedge \star (\star 1 \wedge \star (\star e^j \wedge w)) = b_i e^i$
TODO I bet that could look more pretty with interiors of the volume form.
Isolate the 2-form:
$\sigma \delta_{ij} e^i \wedge \star (\star 1 \wedge \star (\star e^j \wedge \star w)) = c_i e^i$
How to isolate the unique components of any combination of forms in $\Lambda_3$:
TODO Ok I'm getting sick of the vector basis, TODO rewrite this whole worksheet as forms.
$w = a + b_i e^i + \frac{1}{2} c_i \epsilon^{ijk} e^j \wedge e^k + \frac{1}{6} d \epsilon^{ijk} e^i \wedge e^j \wedge e^k$
$w = a + b_i e^i + c_i \star e^i + d \star 1$
$\star w$
$= \star (a + b_i e^i + c_i \star e^i + d \star 1)$
$= (a \star 1 + b_i \star e^i + c_i \star \star e^i + d \star \star 1)$
$= \sigma d + \sigma c_i e^i + b_i \star e^i + a \star 1$
Ok another problem with 0-form + 1-form is that $a \wedge \star b$ is not an inner product anymore.
$(a_0 + a_i e^i) \wedge \star (b_0 + b_j e^j)$
$= a_0 b_0 \wedge \star 1
+ a_i b_0 e^i \wedge \star 1
+ a_0 b_j \wedge \star e^j
+ a_i b_j e^i \wedge \star e^j
$
$= a_0 b_0 \star 1
+ a_0 b_i \star e^i
+ a_i b_j g^{ij} \star 1
$
Notice that unlike the $⋀^1_4$ case of $a \wedge \star b = \star (a \cdot b)$, for the $⋀^0_3 \oplus ⋀^1_3$ case we get extra terms.
Second try, Quaternions as 1-forms in R4 with a +--- signature
Quaternion multiplication table:
$e_\mu \cdot e_\nu =$
$\begin{array}{c|cccc}
& e_0 & e_1 & e_2 & e_3 \\
\hline
e_0 & e_0 & e_1 & e_2 & e_3 \\
e_1 & e_1 &-e_0 & e_3 &-e_2 \\
e_2 & e_2 &-e_3 &-e_0 & e_1 \\
e_3 & e_3 & e_2 &-e_1 &-e_0 \\
\end{array}$
$e_\mu \cdot e_\nu = 2 \delta_{0 ( \mu} e_{\nu)} - \delta_{\mu\nu} e_0 + \epsilon_{0 \mu \nu \rho} e_\rho$
Let $q = q_\mu e^\mu = q_0 e^0 + q_i e^i$
Let $g_{\mu\nu} = diag(1, -1, -1, -1)$ be our metric.
If it were our metric then we could say:
$q^\sharp = q^\mu g_{\mu\nu} e^\nu = q^0 e^0 - q^i e^i$
...and then maybe we could say, for conjugation, that $\bar{q} = q^\sharp$
Maybe we can do quaternion mul as wedge?
$a \wedge b = (a_0 e^0 + a_i e^i) \wedge (b_0 e^0 + b_j e^j)$
$= a_0 e^0 \wedge b_0 e^0
+ a_0 e^0 \wedge b_j e^j
+ a_i e^i \wedge b_0 e^0
+ a_i e^i \wedge b_j e^j
$
$= (a_0 b_i - a_i b_0 ) e^0 \wedge e^i
+ \frac{1}{2} a_i b_j {\epsilon^{ij}}_k \star (e^0 \wedge e^k)$
$a \wedge \star b = (a_0 e^0 + a_i e^i) \wedge (b_0 \star e^0 + b_j \star e^j)$
$= (a_0 e^0 + a_i e^i) \wedge (b_0 \star e^0 + b_j \star e^j)$
$= a_0 b_0 e^0 \wedge \star e^0 + a_i b_j e^i \wedge \star e^j$
$\star a \wedge b = a_0 b_0 \star e^0 \wedge e^0 + a_i b_j \star e^i \wedge e^j$
$= -a_0 b_0 e^0 \wedge \star e^0 - a_i b_j e^j \wedge \star e^i$
Combining inner and exterior:
$a \wedge (b + \star b)$
$= (a_0 b_i - a_i b_0 ) e^0 \wedge e^i
+ \frac{1}{2} a_i b_j {\epsilon^{ij}}_k \star (e^0 \wedge e^k)
+ a_0 b_0 e^0 \wedge \star e^0
+ a_i b_j e^i \wedge \star e^j
$
$= (a_0 b_i - a_i b_0 ) e^0 \wedge e^i
+ \frac{1}{2} a_i b_j {\epsilon^{ij}}_k \star (e^0 \wedge e^k)
+ (a_0 b_0 + a_i b_j g^{ij} ) \star 1
$
All the terms are present, it's just a matter of giving them the right sign and basis.
The fact that the $a_0 b_i - a_i b_0$ signs don't match is frustrating.
So if we self-dual both then
$(a + \star a) \wedge (b + \star b)$
$= a \wedge (b + \star b) + \star a \wedge b$
$= (a_0 b_i - a_i b_0 ) e^0 \wedge e^i
+ \frac{1}{2} a_i b_j {\epsilon^{ij}}_k \star (e^0 \wedge e^k)
+ a_0 b_0 e^0 \wedge \star e^0
+ a_i b_j e^i \wedge \star e^j
- a_0 b_0 e^0 \wedge \star e^0
- a_i b_j e^j \wedge \star e^i$
$= (a_0 b_i - a_i b_0 ) e^0 \wedge e^i
+ \frac{1}{2} a_i b_j {\epsilon^{ij}}_k \star (e^0 \wedge e^k)
+ (a_i b_j - a_j b_i) g^{ij} \star 1
$
Isn't so useful because it cancels the $a_0 b_0$ term.
What if we $a^\sharp \wedge (b + \star b)$ ?
What does $a^\sharp \wedge b$ look like?
$a^\sharp \wedge b = (a^0 e^0 - a^i e^i) \wedge (b_0 e^0 + b_j e^j)$
$= a^0 b_i e^0 \wedge e^i
- a^i b_0 e^i \wedge e^0
- a^i b_j e^i \wedge e^j
$
$= (a^0 b_i + a^i b_0 ) e^0 \wedge e^i
+ \frac{1}{2} a^i b_j {\epsilon^{ij}}_k \star (e^0 \wedge e^k)$
Looks better.
Back