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Is there an inner product on every vector space $V$ (of any infinite dimension) over $\mathbb{R}$ or $\mathbb{C}$? And is there a generalisation of the term "inner product" to vector spaces over any field?

Primarily I need this to prove that if $W$ is a subspace of $V$, then the factor $V/W$ is isomorphic to the orthogonal complement $W^\perp$ of $W$.

J. W. Tanner
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Juraj Hartman
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    How is $W^\perp$ defined if not in an inner product space? – Alex Ortiz Nov 26 '23 at 23:39
  • The first question is of course a duplicate. For example, see https://math.stackexchange.com/questions/247425/is-there-a-vector-space-that-cannot-be-an-inner-product-space (search function FTW). – Martin Brandenburg Nov 26 '23 at 23:49
  • You don't need inner products. Simply prove that $V/W$ is isomorphic to any complement of $W$ (this works over any field). – Anne Bauval Nov 26 '23 at 23:49
  • If the vector space has a dimension, i.e. if it has a basis $\mathcal B$, then there is an isomorphism $\phi:V\to \Bbb C^{\oplus\mathcal B}$, and you can define the inner product $$\langle x,y\rangle =\sum_{b\in \mathcal B} \overline{\phi(x)_b}\phi(y)_b$$ Notice that for all pairs $(x,y)$ the sum on the RHS has finitely many non-zero terms. With ZFC this is always the case. Mind however that (in ZFC) this inner product is complete if and only if $\dim V<\aleph_0$. – Sassatelli Giulio Nov 26 '23 at 23:53
  • @SassatelliGiulio This (except maybe your last sentence) was already in the linked post. – Anne Bauval Nov 26 '23 at 23:55
  • @AnneBauval So? And for what it's worth, it's possible that my last sentence is false. – Sassatelli Giulio Nov 26 '23 at 23:57
  • @JurajHartman In an inner product space, without additionnal hypothesis, $V/W$ is not isomorphic to $W^\perp.$ Take for instance for $W$ a dense hyperplane. – Anne Bauval Nov 27 '23 at 00:06
  • @AlexOrtiz Sorry, I wrote it awkwardly. When I write $W^\perp$, I suppose that there is an inner product on $V$ (and this is the reason I need it). – Juraj Hartman Nov 27 '23 at 00:16
  • @MartinBrandenburg I didn't know about it, thank you for the link. – Juraj Hartman Nov 27 '23 at 00:18
  • @AnneBauval Yes, we need the assumption that every vector in $V$ has the orthogonal projection on $W^\perp$, I forgot to write it. – Juraj Hartman Nov 27 '23 at 00:22
  • But given $V$ and $W$, there generally exists no inner product satisfying this additional condition, and as mentionned in my first comment, you don't need such a product anyway. – Anne Bauval Nov 27 '23 at 07:22

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