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Suppose I have a real vector space $V$ and I would like to extend the scalar multiplication in such a way that I obtain a complex vector space. It is not difficult to see that doing so is equivalent to fixing some linear map $U : V \to V$ satisfying $U^2 = -1$ and then defining $(a + ib)x = ax + bUx$ for $a+bi \in \mathbb{C}$ and $x \in V$. The next natural question is "when does such a map $U$ exist"? The answer is "always", provided $V$ is even-dimensional or infinite dimensional. To see this, take a basis for $V$ and split it into two collections of equal cardinality $(x_i)_{i \in I}$ and $(y_i)_{i \in I}$. Then define $Ux_i = y_i$, $Uy_i = -x_i$ and extend linearly.

But what if I have a real-Banach space $X$? Can I extend the scalar multiplication while ensuring that $X$ becomes a complex Banach space under the original norm? Clearly I need to take $U$ to be an isometry. More specifically, what we'd need is a $U$ satisfying $$ \|ax + bUx\|^2 = (a^2 + b^2) \|x\|^2 $$ for all $a,b \in \mathbb{R}$, $x \in X$.

I suppose this is possible if $X$ is some real Hilbert space $H$ (of even or infinite dimension). One just uses the same trick as in the algebraic case with an orthonormal basis instead instead of a Hamel basis.

Are there any other nontrivial sufficient (or necessary) conditions for a real Banach space to be a complex Banach space in which we have forgotten how to multiply by imaginary scalars?

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In the real Banach space $c_0$, for example, this can't be done: for any $x$ and $y$ in $c_0$ , $\|x + t y\|$ is a piecewise linear function of $t$. –  Robert Israel Nov 11 '11 at 2:03
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up vote 6 down vote accepted

There are many cases where this fails if one insists upon on keeping the original norm (i.e., the isometric version of this problem). A clearly necessary condition for the space to have a complex structure isometrically is that the group of real-linear isometries is uncountable, in particular that it contains a subgroup isomorphic to the multiplicative group of complex numbers of modulus $1$. On the other hand, the/one reason complexification under the original norm fails so often is that every real Banach space can be equivalently renormed so that its group of real-linear isometries is just the two element group $\{ Id_X,\, -Id_X \}$; this fact is due to Jarosz, in Any Banach space has an equivalent norm with trivial isometries, Israel J. Math, 64 (1988), 49-56 (obviously, no renorming is needed in the elementary case $X=\mathbb{R}$!). Others have since studied further what groups can be represented as the group of linear isometries on a Banach space; see, for example, the work of Ferenczi and Galego.

I will also mention that there is a notion of an extremely non-complex Banach space; not only do such Banach spaces fail to admit an operator $U$ satisfying $U^2=-Id$, but in fact every operator $T$ on such a space satisfies $\Vert Id+T^2\Vert = 1+\Vert T^2\Vert$. Properties and the existence of such spaces can be found in the joint work of Koszmider, Martin and Meri.

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Thank you for the interesting references! –  Mike F Nov 11 '11 at 21:40
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