The super group $GL(1|1)$

Question

It is difficult to find information on super groups and I have built my knowledge from various sources. I have the following questions.

$GL(1|1)$ is defined as the group of invertible linear transformations on $\mathbb{C}^{1|1}$.

I see conflicting information on whether the group action should preserve the grading. Is this true? I do not think so, as this would make the matrix diagonal for $GL(1|1)$

Now I can write as a usual $2\times 2$ matrix. What are its entries? Some sources use complex numbers, others use Grassmann variables. Which are supposed to be used?

Regarding invertability, is it true that an arbitrary linear operator on $\mathbb{C}^{1|1}$ is invertible if and only if the $(1,1)$ and $(2,2)$ components of the corresponding matrix are invertible? For example, is it possible to have a matrix of the following type in $GL(1|1)$: $$ \left(\begin{array}{cc} a & b \\ c & 0 \end{array}\right)$$

Answer 1

$GL_{\mathbb{C}}{(1|1)}$ is a $\mathbb{C}$-functor which associates to each super $\mathbb{C}$-algebra $R$
the group of $Aut^{sup}_{R}(\mathbb{C}^{1|1}\otimes_\mathbb{C}R)$ of all super $R$-linear automorphisms on $\mathbb{C}^{1|1}\otimes_\mathbb{C}R$.

$GL_{\mathbb{C}}{(1|1)}(\mathbb{C}) = \Bigl\{\left(\begin{array}{ll}a&0\\0&d\end{array}\right)\Big|a,d \in \mathbb{C}^{\times}\Bigr\}$

$GL_{\mathbb{C}}{(1|1)}(\mathbb{C}^{1|1}) = \Bigl\{\left(\begin{array}{ll}a&b\\c&d\end{array}\right)\Big|a,d \in (\mathbb{C}^{1|1}_{ev})^{\times}\space b,c \in \mathbb{C}^{1|1}_{odd}\Bigr\}$

Answer 2

I see conflicting information on whether the group action should preserve the grading. Is this true? I do not think so, as this would make the matrix diagonal for (1|1)

The group acting should always preserve the grading.

The issue that you are having is that you are considering only the $\mathbb{C}$-points of $\operatorname{GL}(1|1)$ i.e you have only been looking at $\operatorname{GL}(1|1)(\mathbb{C})$ which is, as you have noticed, diagonal.

Since $\operatorname{GL}(1,1)$ is a superscheme (here a Lie supergroup) you need to describe it using the functor of points as described in the answer above.

This will get rid of your problem of wrongly thinking that $\operatorname{GL}(1,1)$ is diagonal.

is it true that an arbitrary linear operator on ℂ1|1 is invertible if and only if the (1,1) and (2,2) components of the corresponding matrix are invertible?

Yes!