2

Let $V$ be a vector space over $\mathbb{C}$. Clifford algebra $Cl(V,Q)$ is defined by the tensor algebra $T^{\boldsymbol{^*}}(V)$ modulo the ideal $I$ generated by $v\otimes v -Q(v,v)\cdot1$.

Let $Cl^{\times}(V,Q)=\{\varphi\in Cl(V,Q): \exists \varphi^{-1} , \varphi\varphi^{-1}=\varphi^{-1}\varphi=1 \}$ be the multiplicative group of units. It is a Lie group by identifying $Cl^{\times}(V,Q)\cong GL(Cl(V,Q))$.(here it is not correct, see (Group of units of a Clifford algebra)

Define the Conjugation of Clifford algebra. \begin{align*} *:v_1\cdot\ldots \cdot v_r\mapsto(-1)^r v_r\cdot\ldots\cdot v_1 \end{align*} Then Spin group is defined by \begin{align*} \text{Spin}(Q)=\{x\in Cl^0: x\cdot x^*=1, x\cdot v\cdot x^*\in V\} \end{align*}

If the multiplication and conjugation are continuous, then $\text{Spin}(Q)$ is closed in $Cl^{\times}(V,Q)$.

Is this correct? If so, why the multiplication and conjugation are continuous?

Jino
  • 111
  • What are your thoughts? In your studies, have you come across similar but maybe easier groups where you have shown analogue results? – Torsten Schoeneberg Nov 07 '22 at 17:19
  • For matrix group, like $O(n)$ is closed in $GL_n(V)$. So does linearity means continuous? – Jino Nov 08 '22 at 01:55
  • 1
    Linear maps on finite-dimensional vector spaces are automatically continuous, see https://en.wikipedia.org/wiki/Discontinuous_linear_map#A_linear_map_from_a_finite-dimensional_space_is_always_continuous, or https://math.stackexchange.com/q/112985/96384 But watch out what is meant by multiplication. It might actually be a bilinear map. And do you understand that the strategy is to use that $Spin(Q)$ is the intersection of two preimages of "obviously" closed sets under continuous maps? – Torsten Schoeneberg Nov 08 '22 at 18:17
  • we have two maps: $\varphi_1:x\mapsto x\cdot x^*$ ,$\varphi_2:(x,v)↦x\cdot v \cdot x$. The two maps are continuous and 1 and V are closed in Cl(V,Q). Their preimage are closed, their intersection is closed. – Jino Nov 09 '22 at 01:33
  • 1
    That looks good. As said, it might just be that you need an argument why $\varphi_2$ is continuous. Note it is not exactly a linear map. – Torsten Schoeneberg Nov 09 '22 at 17:42

0 Answers0