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I'm implementing a Geometric Algebra library where I hope to be able to elegantly handle degenerate metrics as well as the more common euclidean and non-euclidean metrics. i.e. $\ e_i e_i = \{0,1,-1\}$

For $\ e_i e_i = \{1,-1\}$ the most common formula I've found for the dual is $ B^* = BI^{-1} $ where $ I^{-1} $ is the inverse of the psuedoscalar for the space you are working in.

From what I can tell, I need the dual to map a k-vector to a k-vector representing (spanning) it's orthogonal complement. A property to be preserved is $B^{**} = -B$

Which brings up my first question. For what reason does the dual of the dual negate the k-vector it acts on? Presumably you could have $B^{**} = B$ and also still have $B^*\perp B $

Back to the degenerate metric. I can't use the inverse of the psuedoscalar involving a degenerate metric for example: $e_1 e_1 = 0$, $I = e_{123}$ in $ G^{2,0,1}$ since $I^{-1}$ doesn't exist in this case.

If I were to use $B^* = BI$ or $B^* = B \rfloor I$ as I've seen in some material. I end up with $e_1e_{123} = 0$ which is not the orthogonal complement of $e_1$

So what can I do? Best I've come up with is to use difference of the indices against the psuedoscalar's to create an orthogonal k-vector. i.e. $ e_{13} => [1,3]$ diff $[1,2,3] => [2] => e_2$ I've played around with ways of getting the sign to agree with the other definitions of the dual, but can't quite get it right.

Charles Gunn uses Poincare Duality for 3D Projective Geometric Algebra https://bivector.net/doc.html, but this doesn't fully agree with the purely euclidean dual from some quick tests I did. Also I know very little about algebraic topology so implementing a general Poincare Map is beyond me at this point.

Michael Alexander Ewert
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  • It sounds like you want to generalize the $90^\circ$ rotation in the plane. If that's right, it might be done with general metric, but it could be hardly extended to 3 or more dimensions. – Berci Sep 26 '19 at 23:07
  • With $(xe_1+ye_2+ze_3)^2=y^2+z^2$, the null vector $e_1$ is orthogonal to everything; there's no special bivector orthogonal to it. – mr_e_man Sep 26 '19 at 23:24
  • I think you mean "map a $k$-vector to an $(n-k)$-vector". – mr_e_man Sep 26 '19 at 23:28
  • You seem a bit turned around on some of the fundamentals here. From what I can tell, it seems that a review of chapters 1 and 2 of Linear Algebra via Exterior Products by Winitzki might help clear some of the confusion. – Jonathan Trousdale Sep 26 '19 at 23:32
  • @ChainedSymmetry - Exterior algebra has nothing to say about quadratic forms/"metrics"/"dot products". – mr_e_man Sep 26 '19 at 23:37
  • @mr_e_man The post references k-vectors, that's commonly used to denote tensors from the space $\wedge^k V$. – Jonathan Trousdale Sep 26 '19 at 23:42
  • @ChainedSymmetry - These are $k$-vectors in a Clifford algebra. And the exterior product can be defined in terms of Clifford's geometric product. – mr_e_man Sep 26 '19 at 23:43

1 Answers1

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In general, in a geometric algebra over a vector space $V$ with a degenerate metric, there is no unique dual. This is because some vectors are orthogonal to everything (by definition of degenerate); the orthogonal complement is $n$-dimensional, while a dual would give something $(n-1)$-dimensional.

One solution is to simply accept that the dual (defined as contraction with $e_1\wedge e_2\wedge\cdots\wedge e_n$) of a non-zero multivector could be zero.

Another solution is to use the dual space $V^*$ (this is a different kind of dual, often defined as the linear functions from $V$ to $\mathbb R$). If $\{e_1,e_2,\cdots,e_n\}$ is a basis for $V$, then there is a basis $\{\varepsilon_1,\varepsilon_2,\cdots,\varepsilon_n\}$ for $V^*$ satisfying

$$\varepsilon_i\cdot e_j=\delta_{ij}$$

where $\delta_{ij}$ is the Kronecker delta. We may consider this as a dot product in the $2n$-dimensional space $V\oplus V^*$. If we have a non-degenerate metric on $V$, then that determines an isomorphism of $V$ with $V^*$, and thus a non-degenerate metric on $V^*$. But if we have a degenerate metric on $V$, then the corresponding map from $V$ to $V^*$ is not invertible, so we don't get a unique metric on $V^*$ (except perhaps $\varepsilon_i\cdot\varepsilon_j=0$). But the metrics on $V$ and $V^*$ are irrelevant for what you call Poincare duality. This is a map between $k$-vectors in $V$ and $(n-k)$-vectors in $V^*$, which is simply a contraction with the $n$-blade for the other space:

$$B=\langle B\rangle_k\in G_k(V)\subset G(V)\subset G(V\oplus V^*)$$

$$B^*=\pm B\,\lrcorner\,(\varepsilon_1\wedge\varepsilon_2\wedge\cdots\wedge\varepsilon_n)$$

$$B^*=\langle B^*\rangle_{n-k}\in G_{n-k}(V^*)\subset G(V^*)\subset G(V\oplus V^*)$$

$$B=\pm B^*\,\lrcorner\,(e_1\wedge e_2\wedge\cdots\wedge e_n)$$

This contraction can be calculated by repeatedly applying the identities $(A\wedge B)\,\lrcorner\, C=A\,\lrcorner\,(B\,\lrcorner\, C)$ and $a\,\lrcorner\,(B\wedge C)=(a\,\lrcorner\, B)\wedge C+(-1)^kB\wedge(a\,\lrcorner\, C)$ where $a=\langle a\rangle_1$ and $B=\langle B\rangle_k$.

(Also note that $(\varepsilon_n\wedge\cdots\wedge\varepsilon_1)\,\lrcorner\,(e_1\wedge\cdots\wedge e_n)=\det[\varepsilon_i\cdot e_j]=1$, which is related to Gramian matrices.)

mr_e_man
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  • Thanks. I'll need to parse this carefully but seems very much in line with the current scribblings I have on the page in front of me. – Michael Alexander Ewert Sep 27 '19 at 01:26
  • This works really well. Simplification: $e_i = \epsilon_i$ if the metric is non-degenerate. When dualizing a blade I use the familiar formula but using a pseudoscalar where I've replaced all occurrences of the degenerate basis with it's corresponding dual space basis. Then you swap dual basis back for regular basis in the result. – Michael Alexander Ewert Sep 28 '19 at 00:39
  • I think that simplification should be $\varepsilon_i=\pm e_i$ depending on the signature of $e_i$. – mr_e_man Oct 01 '19 at 18:52