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What familiar and intuitive properties of finite dimensional vector spaces fails in infinite dimensions?

For example:

  1. In infinite dimensions there are non-continuous linear maps.

  2. In infinite dimensions always $\dim V <\dim V^*$ and, in particular, $V\not\simeq V^{**}$.

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For normed spaces, compactness of the unit ball springs to mind. – David Mitra Jun 13 '13 at 17:45
Continuous? That implies a normed space, right? Because linear maps are continuous by definition in a topological vector space. – dfeuer Jun 13 '13 at 17:46
up vote 4 down vote accepted

For simplicity, all the vector spaces in the following are over $\mathbb{C}$, or some complete field.

All norms on a finite dimensional vector space are equivalent. This is not true for infinite dimensional vector spaces over (consider $L^p$ norms). I believe this comes from the fact that the unit ball is compact for a finite dimensional normed linear spaces (NLS), but not in infinite dimensional NLS.

The weak topology on a finite dimensional vector space is equivalent to the norm topology. This is always false for infinite dimensional vector spaces. More generally, there are many topologies of interest on an infinite dimensional vector space, but just one of interest on a finite dimensional space (from a linear algebra/functional analysis perspective).

There is a nontrivial translation invariant measure for finite dimensional vector spaces (say over $\mathbb{C} $ or $ \mathbb{R}$, the Lebesgue measure). This is not true for an infinite dimensional Hilbert space (the unit ball has infinitely many disjoint translates of a ball of radius $\sqrt{2}/4$).

This is a property of bounded operators on a NLS: The spectrum of a linear map $T:V \to V$ (the set of $\lambda\in \mathbb{C}$ such that $T-\lambda I$ is not invertible) consists precisely of the eigenvalues of $T$. However, if $V$ is infinite dimensional, then $T-\lambda I$ may not be invertible even if $\lambda$ is not an eigenvalue.

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An infinite dimensional vector space can be isomorphic to one of its proper subspaces ($\mathbb{R}\subset\mathbb{C}$ as vector spaces over $\mathbb{Q}$).

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Why are these isomorphic? Bijections between $\mathbb R$ and $\mathbb C$ are generally ill-behaved. – dfeuer Jun 13 '13 at 19:03
In fact, if $f\colon\mathbb C \to \mathbb R$ is an isomorphism, then it must map $1$ to $m≠0$ and $i$ to $n≠0$. Then $x+iy$ gets mapped to $xm+yn$, and this is clearly not a bijection. – dfeuer Jun 13 '13 at 19:15
@dfeuer: Vector spaces of the same dimension, over the same field are isomorphic. $\dim_{\mathbb Q}\mathbb R=\dim_{\mathbb Q}\mathbb C=\mathfrak c$. – Corvus Jun 13 '13 at 19:21
To expand, take a basis of $\mathbb C$ over $\mathbb Q$ and biject it to a basis of $\mathbb R$ over $\mathbb Q$. We can do this since both have cardinality of the continuum. By linearity, this is a vector space isomorphism. Warning: Requires axiom of choice. – Devlin Mallory Jun 13 '13 at 19:44
Oh, I see where I went wrong now. Silly me. To expand the expansion, let $B$ be a basis for $\mathbb C$. Then $B$ includes as a subset a basis $A$ for $\mathbb R$. $A$ has the cardinality of the continuum for reasons that will be clear to me in a moment, and $|B| \le \mathfrak c$ is obvious. – dfeuer Jun 13 '13 at 19:54

Also if $V$ is finite dimensional vector space, then the algebra $\operatorname{End}(V)$ has no non-trivial $2$-sided ideals. For $V$ infinite-dimensional, $\operatorname{End}(V)$ has proper $2$-sided ideal which consists of endomorphisms of finite rank.

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A linear operator on a finite-dimensional vector space is injective if and only if it is surjective (by the Rank-Nullity Theorem). This is false in infinite-dimensional spaces. For instance consider a space $V$ with a countable basis $(e_i)_{i \in \mathbb{N}}$ and take the forward shift operator $S$, defined by $Se_i = e_{i+1}$. This is injective but not surjective. The backward shift operator $S'$, defined by $S'e_i=e_{i-1}$ (define $e_0=0$) is surjective but not injective.

Another interesting thing to think about is commutators. In a finite-dimensional spaces you can never have operators $P,Q$ satisfying $PQ-QP= \mathrm{id}_V$ (taking the trace gives a contradiction). But this is possible in infinite dimensions. For instance, let $V$ be the polynomial ring $k[x]$ for a field $k$, and let $Pf=\frac{d}{dx}f$ and $Qf=xf$. (See keyword Heisenberg algebra.)

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The second thing you mention is interesting, but I'm wondering how you alter the proof when you are working over a field with characteristic other than 0. – Bryan Mar 3 '14 at 18:46
@Bryan, in nonzero characteristic that second statement is false, and the Heisenberg algebra has nontrivial finite-dimensional representations. – MTS Mar 3 '14 at 20:55

At least they still have a basis, or do they? A basis of $\mathbb R$ as $\mathbb Q$-space cannot be written down explicitly and heavily requires the axiom of choice.

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The statement "every vector space has a basis" is equivalent to the axiom of choice. – dls Jun 13 '13 at 18:21
The person who proved that theorem is sometimes seen around these parts. – dfeuer Jun 13 '13 at 18:27

Assuming the axiom of choice, any vector space admits a basis and so has a norm. Therefore, we may look at topological properties. With this viewpoint, there is the classical result:

Theorem: (Riesz) A normed vector space is locally compact iff it is finite dimensional.

But there are also less known results like:

Property: Let $X$ be a normed vector space and $K \subset X$ be a compact subset. Then $X \backslash K$ is path-connected.

It is clearly false in finite dimension (take a sphere for example).

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