# Schauder bases in Banach spaces with a symmetric $k$-FDD

The Kalton-Peck Banach space $$Z_2$$ (see Section 6 in this paper) does not admit an unconditional basis, but it admits an unconditional, even symmetric, FDD (finite dimensional decomposition) into subspaces of dimension $$2$$, and also admit a Schauder basis which is the union of some natural bases of the $$2$$-dimensional subspaces.

QUESTION: Suppose that, for some $$k\in\mathbb N$$, the Banach space $$X$$ admits a symmetric FDD into subspaces of dimension $$k$$.

Can we assure that $$X$$ admits a Schauder basis?

Yes. If $$(E_n)$$ is a FDD for $$X$$ where each $$E_n$$ has dimension $$k$$, then we can pick a basis $$(e_i^n)_{i=1}^k$$ for each $$E_n$$ with basis constant at most $$\sqrt{k}$$. Then the concatenation of $$(e_i^n)_{i,n}$$ in natural order is a Schauder basis for $$X$$ whose basis constant is less than or equal to $$\sqrt{k}C$$ where $$C$$ is FDD constant. The symmetry is not needed.
I have recently found that there is a reference for the result I needed in Chapter 7 of P.G. Casazza, Approximation properties'', HANDBOOK OF THE GEOMETRY OF BANACH SPACES, VOL. 1. Edited by William B. Johnson and Joram Lindenstrauss. Elsevier, 2001.
Proposition 6.5. Let $$(E_n)$$ be a finite dimensional decomposition (with FDD constant $$K$$) for a Banach space X. If each $$E_n$$ has a basis $$(x^n_i)_{i=1}^{k_n}$$ with basis constant bounded by $$M$$, then $$((x^n_i)_{i=1}^{k_n})_{n=1}^\infty)$$ is a basis for $$X$$ with basis constant $$\leq KM$$.