Any Clifford algebra $\operatorname{Cl}(k, p)$ carries an induced inner product, which is the "trace" on its 0-blade: $\langle AB\rangle_0$ for given elements $A, B$ of the algebra.

This inner product, when restricted to the generating vector space $V$, gives back the inner product on $V$.

Now, my question (maybe entirely trivial, but I could not find it in the standard literature, for instance in Lounesto's great book Clifford algebras and spinors):

What is the interplay of the Clifford product and its induced inner product? Are there any formal laws?

PS. I have done some little googling, and I came up with some references on associative algebras with inner product, but unfortunately there seems to be a certain latitude as far as their definition.

PPS In light of the comments below, I realize that the phrasing of my post is not as clear as it should. Please refer to the comments below and my additional replies for the proper context.

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    $\begingroup$ Would have said the symmetrised "Clifford product" is this "induced inner product" up to a factor of 2 but perhaps I'm misunderstanding one or both of the definitions $\endgroup$ Jun 14, 2020 at 2:41
  • $\begingroup$ @AlexArvanitakis they are not the same, though they are related: you refer to (ab+ ba) /2 which is the symmetric side of the clifford product. I was referring to <ab>_0, which is taking the o-dimensional component of the product. On vectors these two are the same, but not in general $\endgroup$ Jun 14, 2020 at 12:53
  • $\begingroup$ I am afraid your question is not clear (at least the way it is phrased). Do you mean that you are looking for some relation between the $0$-component of the product $A\cdot B$ (for arbitrary elements $A$, $B$) and the bilinear form asssociated to the quadratic form involved in the definition of the Clifford algebra ? $\endgroup$ Jun 14, 2020 at 16:12
  • $\begingroup$ @KonstantinosKanakoglou I agree: my question is not as crystal clear as it should, and I apologize. Let me see if I can make it a bit clearer: you start from a vector space with an inner product (V, Q), you build the clifford algebra. Now, the clifford agebra is an associative algebra PLUS the inner product given by the 0-blade (which of course restrict to the starting Q for V). $\endgroup$ Jun 14, 2020 at 16:50
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    $\begingroup$ @MircoA.Mannucci: Right at the bottom of the Wikipedia entry on the Clifford scalar product: is this what you wanted? The fact that Clifford transposition is the adjoint under the scalar product? $\endgroup$
    – Alex M.
    Jun 22, 2022 at 5:26

1 Answer 1


$ \newcommand\lcontr{\,\lrcorner\,} \newcommand\rcontr{\,\llcorner\,} \newcommand\lcontrr{{\rfloor}} \newcommand\rcontrr{{\lfloor}} \newcommand\form[1]{\langle#1\rangle} \newcommand\Ext\bigwedge \newcommand\rev\widetilde $

I hope that this is in the vein of what you were looking for.

In a review (Marcel Riesz’s Work on Clifford Algebras, 1993) of Marcel Riesz's lecture notes (Clifford Numbers and Spinors, 1958) on Clifford algebras, Pertti Lounesto demonstrates a natural linear isomorphism between $\Ext(V)$ and $Cl(V)$.

Given a symmetric bilinear form $\form{\cdot,\cdot}$ on a vector space $V$ over a field $K$ with characteristic $\not=2$, this extends to a bilinear form on $\Ext V$. On $k$-blades (i.e. simple $k$-vectors), we define $$ \form{x_1\wedge\cdots\wedge x_k, y_1\wedge\cdots\wedge y_k} = \det\form{x_i, y_j}, $$ which is to say that we take the determinant over the matrix with entries $a_{ij} = \form{x_i, y_j}$. For blades $A, B \in \Ext V$ with different grades we define $\form{A, B} = 0$, and then extend by linearity to the entirety of $\Ext V$. Though maybe not obvious when described like this, this extension of the form on $V$ is natural, see here and here.

With this form on $\Ext V$, the adjoints of the exterior product $$ \form{X \wedge Y, Z} = \form{Y, X \lcontr Z},\quad \form{X \wedge Y, Z} = \form{X, Z \rcontr Y} $$ (where $X, Y, Z \in \Ext V$ are arbitrary multivectors) are found to be the left ($\lcontr$) and right ($\rcontr$) contractions of the Clifford algebra defined by $\form{\cdot,\cdot}$ on $V$. They are indispensable when studying Clifford algebras for applications. There are various different conventions that can be used when defining them; see the appendix of this preprint (Compendium on Multivector Contractions, 2022) by André Mandolesi. It's worth noting that the alternative contractions $$ X\lcontrr Y = \rev X\lcontr Y,\quad X\rcontrr Y = X\rcontr\rev Y, $$ where $\rev X$ is the reverse of $X$, are particularly popular; see Lounesto or this article (The Inner Products of Geometric Algebra, 2002) by Leo Dorst for an exposition on their basic properties.

We may note that $a \lcontr b = a \rcontr b = \form{a, b}$ for $a, b \in V$. Now define the products $$ aX = a\lcontr X + a\wedge X,\quad Xa = X\rcontr a + X\wedge a, $$ which can be seen to be exactly the Clifford algebra products of $a$ and $X$. By this definition, this product is self-adjoint: $\form{aX, Y} = \form{X, aY}$ with an analogous formula for $a$ on the right. But any blade $A$ can be written as $A = a_1a_2\cdots a_k$ for some $k$ and $a_1,\dotsc, a_k \in V$ using Clifford products (which are associative); hence $$ \form{AX, Y} = \form{a_1a_2\cdots a_kX, Y} = \form{a_2a_3\cdots a_kX, a_1Y} = \form{a_3a_4\cdots a_kX, a_2a_1Y} = \form{X, \rev AY}. $$ By linearity, the same holds for when $A$ is an arbitrary multivector. Thus, $$ \form{X, Y} = \form{X\cdot 1, Y} = \form{1, \rev XY} = \form{\rev XY}_0. $$

It's worth noting that the contractions have a very direct connection with the Clifford product. For any $X, Y \in \Ext V$, $$ X\lcontr Y = \sum_{j=0}^n\sum_{k=0}^n\form{\form{\rev X}_j\form{Y}_k}_{k-j},\quad X\rcontr Y = \sum_{j=0}^n\sum_{k=0}^n\form{\form{X}_j\form{\rev Y}_k}_{j-k}, $$ where $\form{\cdot}_l = 0$ if $l < 0$. For $a \in V$ and $X \in \Ext V$ in particular, $$ a\lcontr X = \frac12(aX - \hat Xa),\quad X\rcontr a = \frac12(Xa - a\hat X), $$ where $\hat X$ is the grade involution (i.e. main involution) applied to $X$.



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