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# Is this topology basis dependent?

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Consider a topological field $K$ and an algebraic(!) vector space $V$ over $K$, that is, $V$ has not (yet) a topology defined on it. I'm particularly interested in the case where $V$ has infinite dimension.

Now be $V^\ast$ the algebraic dual of $V$. Define a topology on $V^\ast$ through pointwise convergence, or equivalently, consider $V^\ast$ as subset under the subspace topology of $K^V$ under the product topology. This topology obviously makes $V^\ast$ a topological vector space over $K$.

Denote with $V^{[\ast]}$ the subspace of $V^\ast$ which consists of all $\phi\in V^\ast$ whose kernel has finite codimension, that is $V/\operatorname{ker}(\phi)$ has finite dimension. Obviously $V^{[\ast]}$ also is a topological vector space over $K$.

Now consider an arbitrary basis $\{b_i:i\in I\}$ of $V$ (where $I$ is an appropriate index set). Then one can define covectors $\{\omega_i:i\in I\}$ by $\omega_i(b_j)=\delta_{ij}$. Of course different bases ${b_i}$ lead to different covectors $\omega_i$.

However (if I made no error in my thoughts) in all cases $\{\omega_i:i\in I\}$ is a basis of $V^{[\ast]}$. Therefore the linear map $f:V\to V^{[\ast]}$ which maps $b_i$ to $\omega_i$ is a vector space isomorphism. Now we can define a set $U\subseteq V$ to be open if its image under $f$ is open. This topology then also makes $V$ a topological vector space over $K$.

My question now is:

Does this topology on $V$ depend on the choice of basis $\{b_i\}$?

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## 1 answer

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First of all, the kernel of any $\phi\in V^\ast$ has codimension at most one; more precisely, $\ker\phi = V \Leftrightarrow \phi = 0$ and $\operatorname{codim}\ker\phi = 1$ if equalities are false. Hence $V^{[\ast]} = V^\ast$. Ī̲ suspect that by $V^{[\ast]}$ you really assume a basis-dependent thing, namely finite linear combinations of ${\omega_i}$ mentioned below. Such $\psi\in V^\ast$ that $\forall i\in I: \psi(b_i) = 1$ cannot be expressed in omegas unless $I$ is finite, but (looking at algebra alone) $\psi$ is not essentially different from any non-zero covector.

Secondly, your ${\omega_i}$ obviously are linearly independent. Although if $I$ is infinite, then ${\omega_i:i\in I}$ will not constitute a basis of $V^\ast$ (as explained above), your $f$ is anyway a monomorphism and you can pull the topology back to $V$ this way.

Thirdly, the topology is basis-dependent unless the dimension is finite. It follows from the fact that a hyperplane $\ker\phi$ (where $\phi\in V^\ast$) is closed* in $V$ if and only if $\phi$ is generated by ${\omega_i:i\in I}$; see the first paragraph again. Note that Ī̲ require {0} to be closed in $K$ to disqualify trivial topology.

* With my intuition it would be more convenient to check half-spaces: whether is $\phi > 0$ open and $\phi \le 0$ is closed. But $K$ isn’t an ordered field, alas.

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#### 1 comment

Thank you for your answer (and sorry for my late reaction). Indeed I was after the space spanned by finite linear combinations of the $\omega_i$; I thought (incorrectly, as I now see) that I had found a base-independent characterisation, which would have meant that it is base independent. So IIUC, the topology is base dependent because the spanned subspace in $V^*$ is? So if I have two bases whose $\omega_i$ span the same subspace, the induced topology for those two bases is the same? celtschk‭ about 1 month ago This community is part of the Codidact network. We have other communities too — take a look!

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