# $A\oplus B\cong A\oplus C$ implies $B\cong C$? (No, it does not)

I am asked to prove that for $p\in (1, \infty)$, $$L_{p}[0,1]\cong L_{p}[0,1]\oplus \ell_{2}$$ on a homework assignment, and I think I can show using results from class that $\ell_2\oplus \ell_2\cong \ell_2$.

From this I could say that $L_{p}[0,1]\oplus \ell_{2}\cong L_{p}[0,1]\oplus \ell_{2}\oplus\ell_{2}$

From here I feel like I should be able to conclude that $$L_{p}[0,1]\cong L_{p}[0,1]\oplus \ell_{2}$$

But I know of no such result that allows me to do this. Can anyone tell me if it's true or false?

EDIT: Definitely false. (See counter example below from Arturo Magidin).

That is, if $B\oplus A\cong C\oplus A$ can I conclude that $B\cong C$?

Proper Solution (based on hints below from t.b.):

1) Prove that $\ell_2\oplus \ell_2\cong \ell_2$

2) Use the fact that $\ell_2$ is complemented in $L_p[0,1]$ to write $L_p[0,1] = \ell_2\oplus (\ell_2)^{c}$.

3) Then I combine these to obtain:

$L_p[0,1]\cong \ell_2\oplus (\ell_2)^{c}\cong (\ell_2\oplus \ell_2) \oplus (\ell_2)^{c} \cong \ell_2 \oplus (\ell_2\oplus (\ell_2)^{c})\cong \ell_2\oplus L_p[0,1]$.

I skipped some pieces of your more general argument. I was just wondering if I did anything illegal, so to speak.

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You say that you can prove that $\ell_2\oplus\ell_2\cong\ell_2$. Note that this means that you can prove that $\ell_2\oplus\ell_2\cong \ell_2\oplus \mathbf{0}$. If the statement you want were true, then you would be able to conclude that $\ell_2\cong \mathbf{0}$. Is that true? – Arturo Magidin Mar 19 '12 at 1:38
Wow. Thanks for putting it so simply. I had tried many examples and they all seemed to work. But I never thought of trying with that. Thanks! I feel silly now. – roo Mar 19 '12 at 1:40
@t.b. Sorry I got confused between the two spaces being equal and them being isomorphic. – user38268 Mar 19 '12 at 2:24
Wouldn't this work: $$L_p\cong \ell_2\oplus Y\cong ( \ell_2\oplus\ell_2)\oplus Y\cong \ell_2\oplus(\ell_2 \oplus Y)\cong \ell_2\oplus L_p,$$ or, is this just t.b.'s argument? (it's late for me...) – David Mitra Mar 19 '12 at 2:27
To address your general question about whether direct decompositions are unique (i.e. whether $B \cong C$ in your case), you might also consider the very small/simple counter-example of Bjarni Jonsson described here.. This is discussed at length in Algebras, Lattices, Varieties chapter 5. – William DeMeo Mar 19 '12 at 2:35

Since Arturo explained why your idea doesn't work, here's a

Hint:

Let $X = L^p$ and let $Y = L^p \oplus \ell^2$.

Prove:

1. $X^2 \cong X$ and $Y^2 \cong Y$.

2. $X$ is isomorphic to a complemented subspace of $Y$ and vice versa (you probably know that $L^p$ has a complemented subspace isomorphic to $\ell^2$, see e.g. Proposition 6.4.2 in Albiac-Kalton or follow David Mitra's suggestion, Corollary 9.2 on page 90 in Carother's A Short Course on Banach Space Theory).

3. Now you are in position to apply Pełczyński's argument to conclude that $X \cong Y$ (see point 3. of my question here for the details).

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Ahhhhhh! I love it. Thanks. The $X^2\cong X$ I hadn't thought of doing, although I literally did it for $X:=C[0,1]$ when I was trying to test my original (but false) claim. Thanks again! – roo Mar 19 '12 at 1:56
I included my argument into my original question. I didn't really follow your process exactly, but just used the argument in your referenced post about $L_\infty$. Does my argument still work? Or am I misunderstanding a key concept? edit: (just saw you comment above. Thanks very much again!!!) – roo Mar 19 '12 at 2:25
That's why I wrote: yes, that's it! (that was directed at you) :) – t.b. Mar 19 '12 at 2:26