Let $A$ be a $C^*$-algebra. A sub space $I$ of $A$ is called Lie ideal of A if $[I,A]= IA-AI \subset I$
Since I contains $0$, isn’t it this definition equivalent to definition of two sided ideal of $C^*$-algerba?
Most probably I’m missing something in the definition of Lie ideal. Any ideas?
First intended this as a response to a comment, but answering in comments is silly. I'm basically just giving a bit more detail to Dietrich's comment.
The product is closed for $A$, but not necessarily for an arbitrary subspace $I$. So think for example $A = \mathfrak{gl}(n)$ the C*-algebra of all $n \times n$-matrices, and $I = \mathfrak{sl}(n)$ the subspace of matrices with trace zero. Every commutator $XY - YX$ has trace zero, so $IA - AI \subset I$, but the product $XY$ of a matrix with a trace-zero matrix does not necessarily have trace zero again, e.g.
$$\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}. $$
Hence $IA$ and $AI$ are not in $I$. So $I$ is a Lie idal in $A$, but not an ideal with respect to multiplication.
I think the confusion might come from the notation $IA - AI$. This should probably be read as the subspace spanned by $\{ ia - ai : i \in I, a \in A\}$, and not by $\{ia - bj : i,j \in I, a,b \in A\}$. If it was the latter, I believe your interpretation would be correct, but then the space would lose its "Lie-algebraic" character.