# C*-algebra with certain involution and multiplication is an *-algebra

• HeinzBor
In summary, the conversation discusses the properties of a $C^*$-algebra and its unital extension. The unital extension, denoted by $\overline{\mathcal{A}}$, is shown to be a $*$-algebra with the unit $1 = (0,1)$. The difficulties in showing this include proving that the unit is indeed a unit and that the involution is preserved. It is also shown that the vector space structure can be used to simplify the calculations. A helpful exercise is suggested to practice using these properties.f

#### HeinzBor

Homework Statement
C*-algebra which may or may not be unital.
Relevant Equations
involution and multiplication given by:
Let $$\mathcal{A}$$ be a $C*$-algebra which may or may not have a unit with norm $$||.||$$, and put $$\mathcal{\overline{A}} = \mathcal{A} \oplus \mathbb{C}$$ as a vector space with mupltiplication:

$$(a, \lambda) (b, \mu) = (ab + \lambda b + \mu a , \lambda \mu)$$,
$$(a, \lambda)^{*} = (a^{*}, \overline{\lambda}).$$

Then I must show that $$\mathcal{\overline{A}}$$ is a $$*$$-algebra with unit 1 = (0,1).

1)
(0,1) is the unit since:
$$(a, \lambda)(0,1) = (a0 + \lambda 0 + 1a , \lambda 1) = (a, \lambda)$$

2)
$$(A^{*})^{*} = A \forall \mathcal{\overline{A}}$$ since:
$$((a, \lambda)^{*})^{*} = (a^{*}, \overline{\lambda})^{*} = (a^{**}, \overline{\overline{\lambda}}) = (a, \lambda)$$.

However I am struggling a bit with showing that

3) $$(aA + bB)^{*} = (\overline{a} A^{*} + \overline{b} B^{*}) \ \forall A,B \in \overline{\mathcal{A}}, a,b \in \mathbb{C}$$

and

4) $$(AB)^{*} = B^{*} A^{*}$$

\begin{align*}
(\mu A +\nu B)^*&=(\mu(a,\alpha )+\nu(b,\beta ))^*=((\mu a,\mu \alpha )+(\nu b,\nu \beta ))^*\\
&=(\mu a+ \nu b,\mu \alpha +\nu \beta )^*=((\mu a+\nu b)^*,\overline{\mu \alpha +\nu \beta})\\
&=(\overline{\mu}a^*+\overline{\nu}b^*,\overline{\mu} \cdot \overline{\alpha }+\overline{\nu}\cdot \overline{\beta })\\
&= \ldots
\end{align*}

\begin{align*}
(\mu A +\nu B)^*&=(\mu(a,\alpha )+\nu(b,\beta ))^*=((\mu a,\mu \alpha )+(\nu b,\nu \beta ))^*\\
&=(\mu a+ \nu b,\mu \alpha +\nu \beta )^*=((\mu a+\nu b)^*,\overline{\mu \alpha +\nu \beta})\\
&=(\overline{\mu}a^*+\overline{\nu}b^*,\overline{\mu} \cdot \overline{\alpha }+\overline{\nu}\cdot \overline{\beta })\\
&= \ldots
\end{align*}

Alright now it is clear! I was stuck because I didn't know that I was allowed to set $$(\mu a + vb)^{*}) = (\overline{\mu} a^{*} + \overline{v} b^{*})$$. (which should just be applying the involution) But from your last step it is just one time applications of vector space structure, pull out the scalars and then one substitution thanks a lot!

Last edited:
Alright now it is clear! I was stuck because I didn't know that I was allowed to set $$(\overline{\mu} a^{*} + \overline{v} b^{*}) = ((\overline{\mu} a^{*}, \overline{\mu} \ \overline{\alpha} ))$$. But from your last step it is just one time applications of vector space structure, pull out the scalars and then one substitution thanks a lot!
Typo?

Typo?
big time , fixed now

For the last one I end up here.

$$(b, \beta)^{*} (a, \alpha)^{*} = (b^{*}, \overline{\beta})(a^{*}, \overline{\alpha}) = (b^{*} a^{*} + \overline{\beta}a^{*} + \overline{\alpha}b^{*}, \overline{\beta} \overline{\alpha}) = (a^{*} B^{*} + \overline{\alpha} b^{*}, \overline{\beta} \overline{\alpha})$$

\begin{align*}
(b^*a^*+\overline{\beta }a^*+\overline{\alpha }b^*,\overline{\beta }\cdot \overline{\alpha })&=
(b^*a^*,\overline{\beta }\cdot \overline{\alpha })+(\overline{\beta }a^*,\overline{\beta }\cdot \overline{\alpha })+(\overline{\alpha }b^*,\overline{\beta }\cdot \overline{\alpha })\\
&=(ab,\alpha \beta )^* + \left(\beta (a,\alpha )\right)^*+\left(\alpha (b,\beta )\right)^*\\
&=\left((ab,\alpha \beta ) +\beta (a,\alpha )+\alpha (b,\beta )\right)^*\\
&=\left((a,\alpha )\cdot (b,\beta )\right)^*
\end{align*}

It is all about the definitions you gave in post #1 and the previous linearity.

If you want to practice such things then a good exercise would be the following:
Prove that ##(G,\cdot)## is a group if and only if ##xa=b## and ##ax=b## have unique solutions for all ##a,b\in G.##

What's also helpful: enumerate all properties a ##C^*##-algebra has, extended by those you have already proven for its unital extension, and write the numbers at each step of your calculations. This way you see what you used where.

\begin{align*}
(b^*a^*+\overline{\beta }a^*+\overline{\alpha }b^*,\overline{\beta }\cdot \overline{\alpha })&=
(b^*a^*,\overline{\beta }\cdot \overline{\alpha })+(\overline{\beta }a^*,\overline{\beta }\cdot \overline{\alpha })+(\overline{\alpha }b^*,\overline{\beta }\cdot \overline{\alpha })\\
&=(ab,\alpha \beta )^* + \left(\beta (a,\alpha )\right)^*+\left(\alpha (b,\beta )\right)^*\\
&=\left((ab,\alpha \beta ) +\beta (a,\alpha )+\alpha (b,\beta )\right)^*\\
&=\left((a,\alpha )\cdot (b,\beta )\right)^*
\end{align*}

It is all about the definitions you gave in post #1 and the previous linearity.

If you want to practice such things then a good exercise would be the following:
Prove that ##(G,\cdot)## is a group if and only if ##xa=b## and ##ax=b## have unique solutions for all ##a,b\in G.##

What's also helpful: enumerate all properties a ##C^*##-algebra has, extended by those you have already proven for its unital extension, and write the numbers at each step of your calculations. This way you see what you used where.
Yes starting with the vector space structure axiom, I wasn't sure I could use it in that way in this setting.. Thanks again! I will try to go through that it has been too long ago since I took an algebra course!