# Does tensor product with $L_p$ operator algebra preserve exact sequences?

By $L_p$ operator algebra I mean a closed subalgebra of the algebra of bounded linear operators on some $L_p$ space where $p\in(1,\infty)$. There is a notion of tensor product of $L_p$ spaces (as described, for instance, in Defant & Floret's Tensor Norms and Operator Ideals), which then allows one to talk about tensor products of $L_p$ operator algebras. Given an exact sequence $0\rightarrow I\rightarrow A\stackrel{\pi}{\rightarrow} A/I\rightarrow 0$ of $L_p$ operator algebras, under what condition can I take the tensor product of each term with another $L_p$ operator algebra $B$ and still have an exact sequence? In particular, if I have a linear section $s:A/I\rightarrow A$, i.e., $\pi\circ s=id$ (and assumed to be $p$-completely contractive if necessary), does taking tensor products preserve exactness?

It seems that there is a slight problem here. You want to talk about subalgebras of $B(L_p(\mu))$ for some measure $\mu$ but at the same time you want to work with a tensor product that combines two $L_p(\mu)$-spaces and as a result gives a space of the same form. This does seem suit your purposes.
For example $C(X)$-algebras are $L_p$-operator algebras but I don't think that your tensor product makes them $L_p$-operator (Banach) algebras, even for $p=2$, which corresponds to the Hilbert-space tensor product.
• What I was trying to describe was indeed the spatial $L_p$ tensor product as discussed in 1.14-1.15 of Phillips's paper (arxiv.org/pdf/1309.6406.pdf). Have I confused myself with different notions of tensor product? – cyc Aug 15 '16 at 12:36
• @cyc, I didn't get why do you need any abstract tensor product as you may simply take representations on $L_p(\nu \otimes \mu)$. – Tomek Kania Aug 15 '16 at 12:42