# Geometric algebra: is a $k$-vector simple iff its square is a scalar?

In Geometric Algebra, a multivector $$u$$ is simple if it is of the form $$u = v_1\wedge \cdots\wedge v_k$$ for some vectors $$v_i$$ (a.k.a., $$u$$ is a $$k$$-blade). Clearly, a multivector can only be simple if it a $$k$$-vector, that is, if it is not a mixed multivector.

Is there an easy test of whether a given $$k$$-vector is simple?

‘Easy’ meaning something that could be implemented algorithmically. (This is perhaps related to the problem to diagnosing whether a differential $$p$$-form is simple.)

My conjecture is that a $$k$$-vector $$u$$ is simple iff $$u^2$$ is a scalar, where $$u^2 = uu$$ is the geometric product of $$u$$ with itself. I think I have shown the $$\implies$$ direction (see below), and have not been able to find a counterexample to the $$\Longleftarrow$$ direction: that is, an example of a $$k$$-vector whose geometric-square is a scalar but which is not simple.

Proof that a simple $$k$$-vector has scalar geometric-square:

Let $$u = v_1\wedge\cdots\wedge v_k$$ be any simple $$k$$-vector. We may write \begin{align} u &= v_1\wedge v_2\wedge \cdots v_k \\ &= \underbrace{\left(v_1 - \operatorname{proj}_{v_2}(v_1)\right)}_{\bar v_2}\wedge v_2\wedge\cdots\wedge v_k ,\end{align} where $$\operatorname{proj}_{v_2}(v_1) = \frac{v_1\cdot v_2}{v_2\cdot v_2}v_2$$ since $$\operatorname{proj}_{v_2}(v_1)\wedge v_2 = 0$$. Notice that $$\bar v_1 \cdot v_2 = 0$$ by construction, and therefore $$\bar v_1 \wedge v_2 = \bar v_1 v_2$$ (since $$ab = a\cdot b + a\wedge b$$ for $$1$$-vectors). We proceed inductively, using associativity: \begin{align} u &= \bar v_1 (v_2\wedge v_3) \cdots \wedge v_k \\ &= \bar v_1 \left(v_2 - \operatorname{proj}_{v_3}(v_2)\right)\wedge v_3 \wedge \cdots \wedge v_k \\ &= \bar v_1 \bar v_2 v_3 \wedge \cdots \wedge v_k \\&\ \;\vdots \\ &= \bar v_1\bar v_2 \cdots \bar v_k \end{align} Finally, we normalise $$e_i := \bar v_i/\|\bar v_i\|$$ and introduce an appropriate scaling factor $$\lambda = \|\bar v_1\|\cdots\|\bar v_k\|$$. Notice that $$\{e_i\}$$ is orthonormal.

Thus, $$u$$ is a $$k$$-blade, whose square is trivially a scalar: $$uu = \lambda^2 e_{i_1}\cdots e_{i_k} e_{i_1}\cdots e_{i_k} = \pm \lambda^2 e_{i_1}^2\cdots e_{i_k}^2 = \pm \lambda^2 \|e_{i_1}\|^2\cdots \|e_{i_k}\|^2.$$ (In the middle equality, a factor of $$\pm1$$ is generated from commuting the $$e_i$$’s which satisfy $$e_ie_j = -e_je_i$$ if $$i\ne j$$.)

• In the beginning, you write $v_1\wedge\ldots\wedge v_k$, later on you write $v_1\ldots v_k$, are these two products the same? Commented May 3, 2020 at 8:23
• Are you counting $e_1e_2e_3e_4$ as a scalar? (I think it is sometimes called a pseudo-scalar, but calling it a scalar seems like a stretch.) Commented May 3, 2020 at 8:28
• Given a $k$-vector $v\in \Lambda^kV$ consider the linear map $C_v:\Lambda^{k-1}V\to V$ defined (using dot product) by $(C_vw,x):=(v, w\wedge x)$. Its rank is at least $k$ (unless $v=0$), and a standard result is that $v$ is fully decomposable ("simple") if and only $C_v$ has minimal rank $k$, see Dibag, Factorization in Exterior Algebras, p.262. This is algorithmic since rank computation is algorithmic. Also related, Deciding whether a form in the exterior power $\Lambda^kV$ is decomposable. Commented May 3, 2020 at 9:02

Clarification: In this context, “$$u$$ is simple” means “$$u$$ is a $$k$$-blade” means “$$u = \vec u_1\wedge \cdots \wedge \vec u_k$$ for some vectors $$\vec u_i$$”.

Answer: No. $$u$$ is simple $$\Longrightarrow$$ $$u^2$$ is scalar $$\,\,\not\!\!\Longrightarrow$$ $$u$$ is simple

Proof:

If $$u$$ is simple, then there exists an orthonormal basis for which $$u = λe_1\cdots e_k$$; then $$u^2 = (-1)^{\frac{k(k-1)}2}\tilde{u}u = (-1)^{\frac{k(k-1)}2}λ^2 e_1^2 \cdots e_k^2$$ is a scalar.

A counter-example to the reverse implication is $$u = e_1e_2e_3 + e_4e_5e_6$$. This is a $$3$$-vector which is not simple, since it cannot be factored into $$u = \vec u_1 \wedge \vec u_2 \wedge \vec u_3$$. However, its square is a scalar: $$u^2 = -e_1^2e_2^2e_3^2 - e_4^2e_5^2e_6^2$$ which is $$-2$$ assuming a Euclidean signature $$e_i^2 = 1$$.

Actually even your forward direction is incorrect: it is not true that the square of what you call a simple $$k$$-vector is a scalar (or a pseudo-scalar, or a central element). The mistake in your proof is that you cannot write $$u=\lambda e_{i_1}\cdots e_{i_k}$$ in general.

As a counter-example (assuming $$(e_i)$$ is an orthonormal basis): $$u=e_1\cdot (e_1+e_2)=e_1e_2+1$$ (so $$v_1=e_1$$ and $$v_2=e_1+e_2$$) gives $$u^2=(e_1e_2)^2+2e_1e_2+1=2e_1e_2$$ which is not central or anything if the dimension of the quadratic space is at least $$3$$.

• Though hang on... by simple (a.k.a. decomposable) we mean $u$ is the $\wedge$-product of $k$ vectors (not the geometric product of $k$ vectors). So $u = e_1(e_1 + e_2) = 1 + e_1e_2$ is indeed not simple. I believe I fleshed out my proof that simple $k$-vectors may be written as $k$-blades. Commented May 3, 2020 at 11:57
• The issue is that given a quadratic space $(V,q)$ you can either work in $\bigwedge V$ or in the Clifford algebra $Cl(V,q)$ (or geometric algebra if you want), but there is no natural identification between the two. So it does not really make sense to talk about both products at the same time. If you are working in a special situation that allows that, you should explain it in your question. Commented May 3, 2020 at 12:41
• @CaptainLama - There is a natural identification between the two. The exterior product is the antisymmetric part of the geometric product. See en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/Geometric_algebra#The_geometric_product Commented Jun 9, 2020 at 0:41
• For reference, this answer is wrong. Commented Jan 4, 2023 at 21:33