Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In reading Sternberg's notes on Clifford algebras and spin representations (page 148) I encountered the following:

"...Consider the linear map $$C(\mathbf p)\rightarrow \wedge \mathbf p, x\mapsto x1$$ where $1\in \wedge^{0}\mathbf p$ under the identification of $\wedge^{0}\mathbf p$ with the ground field. The element $x1$ on the extreme right means the image of 1 under the action of $x\in C(\mathbf p)$. "

I am wondering how to derive the explicit form of this map. Sternberg give a homomorphism $C(\mathbf p)\rightarrow \operatorname{End}{(\wedge \mathbf p)}$ by extending the map $\mathbf p\rightarrow \operatorname{End}{(\wedge \mathbf p)}$, which is defined by $v\mapsto \epsilon(v)+\iota(v)$. In here $\epsilon(v)$ denotes the exterior mulplication by $v$ and $\iota(v)$ be the the adjoint of $\epsilon(v)$ relative to the biinear form given by $(x_{1}\wedge\cdots\wedge x_{k},y_{1}\wedge\cdots\wedge y_{k})=\det((x_{i},y_{j}))$.

Hence we have $ \epsilon(v)+\iota(v)(u)=v\wedge u+Au$, with $(v\wedge u, w)=(u,Aw),\forall u,v,w\in \wedge\mathbf p$. Sternberg argues that $(\epsilon(v)+\iota(v))^{2}=(v,v)_{\mathbf p}\operatorname{id}$, therefore we may extend it via universal property to a map $C(\mathbf p)\rightarrow \wedge(\mathbf p)$.

This relationship is not clear to me because I do not see how $\epsilon(v)(\iota(v)(u))+\iota(v)(\epsilon(v)(u))=(v,v)_{\mathbf p}u$, namely LHS is in exterior product while RHS is in a nice closed form.

This confusion hindered me to understand the nature of the linear map he gave. For example, Sternberg wrote: $$v_{1}v_{2}\rightarrow v_{1}\wedge v_{2}+(v_{1},v_{2})1$$ and $$v_{1}v_{2}v_{3}\rightarrow v_{1}\wedge v_{2}\wedge v_{3}+(v_{1},v_{2})v_{3}-(v_{1},v_{3})v_{2}+(v_{2},v_{3})v_{1}$$ I do not know how to derive these formulas explicitly, I believe they should be elementary in nature and easy to work out by hand. So I must have missed something.

share|cite|improve this question
Edited, thanks for pointing out. Really embarrassing mistake. – Kerry Nov 20 '11 at 7:02
Use \wedge ($\wedge$) instead of \vee ($\vee$) for the wedge product. – wildildildlife Nov 20 '11 at 13:11
updated. Thanks. – Kerry Nov 21 '11 at 1:11
You can WLOG assume that $u$ is a "pure wedge", i. e., of the form $u=u_1\wedge u_2\wedge ...\wedge u_n$ for some $u_1,u_2,...,u_n\in\mathbf p$ (because such "pure wedges" generate the vector space $\wedge\left(\mathbf p\right)$). Then both $\epsilon(v)(\iota(v)(u))$ and $\iota(v)(\epsilon(v)(u))$ can be expanded into sums, and you will easily see that the addends of these sums cancel out pairwise except of one which is exactly $(v,v)_{\mathbf p}u$. – darij grinberg Nov 29 '11 at 17:17
Note: Working with $\iota$ becomes much easier once you explicitly write down $\iota$ as follows: $\left(\iota\left(v\right)\right)\left(u_1\wedge u_2\wedge ...\wedge u_n\right) = \sum\limits_{i=1}^n \left(-1\right)^{i-1}\left(v,u_i\right)_{\mathbf p} u_1\wedge u_2\wedge ...\wedge \hat{u_i}\wedge ...\wedge u_n$. – darij grinberg Nov 29 '11 at 17:18
up vote 1 down vote accepted

It is enough to consider action of $\epsilon_v \equiv \epsilon(v)$ and $\iota_v \equiv \iota(v)$, $v \in {\bf p}$ on a basis of $\wedge {\bf p}$.

$\epsilon_v(1) \equiv v$; $\epsilon_v(v_1 \wedge \cdots \wedge v_k) \equiv v \wedge v_1 \wedge \cdots \wedge v_k$.

$\iota_v(1) \equiv 0$; $\iota_v(v_1 \wedge \cdots \wedge v_k) \equiv \sum_{j=1}^k (-1)^{j-1} (v,v_j)_{\bf p} v_1 \wedge \cdots \hat{v}_j \cdots \wedge v_k$,

where $\hat{v}_j$ means that $v_j$ is omitted from product.

Then for any $v, w \in {\bf p}$ and $x = x_1 \wedge \cdots \wedge x_k \in \wedge {\bf p}$

$\iota_v (\epsilon_w (x)) = \iota_v (w \wedge x) = (v,w)_{\bf p} x - w \wedge \iota_v (x)$

So $\iota_v (\epsilon_w (x)) + \epsilon_w (\iota_v (x)) = (v,w)_{\bf p} x$

after omitting $x$ we have $\iota_v \epsilon_w + \epsilon_w \iota_v = (v,w)_{\bf p} {\rm id}$.

E.g. see Gilbert J.E., Murray M.A.M. Clifford algebras and Dirac operators in harmonic analysis (CUP, 1991)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.