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I think my question is related to Free vector space over a set, Non-numerical vector space examples and Formal (series/sum/derivative…).

I'm reading a paper, which starts with the following sentence (the original source is "Flag Algebras" by Razborov)

Let $\mathbb{R}F$ be the real vector space with basis comprised of all elements of $F$; in other words, the space of all formal, finite $\mathbb{R}$-linear combinations of elements of $F$.

Let me summarize my understandings first. We can define the set $\mathbb{R}F$ as a formal sum as $\mathbb{R}F=\{\sum_{f \in F} \lambda_f f | \lambda_f \in \mathbb{R}\}$ and construct the structure of a vector space as described in Free vector space over a set.

Now, here is my trouble: If $\mathbb{R}F$ is supposed to be a vector space, then, due to the vector space axioms, the following must be satisfied: Let $a, b \in F$, then $a+b=c \in F, \forall a,b \in F$, without defining the actual operation "+" of our vector space (since we speak of "formal" sum here). But if $F$ is the basis of our vector space, then all elements of $F$ must be linear independent, thus there cannot exist a $c \in F$ for which $a+b=c$ with $a,b \in F$. (my definition for a vector space is taken from "Advanced Linear Algebra" by Steven Roman").

I would suppose that we define the vector space as the set of all formal sums of elements of $F$ with real coefficients and take the axioms for granted. This works fine for associativity, the inverse element, etc., since we do not define "+" and "juxtaposition" anyway, but I don't see how the basis can be the whole set $F$.

My question is, how can the vector space axioms be satisfied? Can someone give me some examples, or clarifications?

Ubuntix
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    You cannot add elements of $F$. $\mathbb{R}F$ is the vector space, not $F$. So you don't want to write $a+b=c\in F$ but rather $(\sum_i \lambda_i f_i) ,,+ ,,(\sum_j \lambda_j f_j) = \ldots$ (another formal sum). – Peter Franek Nov 15 '20 at 21:50
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    Usually a way to construct this is to take $\mathbb{R}F$ as the set of all finitely supported functions $f\colon F \to \mathbb{R}$. Realising it as a subspace of an existing vector space (one still has the prove that the “finitely supported” property is preserved under addition and scalar multiplication) shortcuts around having to check a lot of axioms. Then expressions like $a + 2b$ for $a, b \in F$ can be understood just as shorthand notation for these finitely supported functions. – Joppy Nov 15 '20 at 21:54

2 Answers2

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You did not apply the vector space axioms correctly. If $\mathbb{R} F$ is a vector space, then for all $a,b \in \mathbb{R} F$ (not $F$, as you stated), we have some $c \in \mathbb{R} F$ (not $F$) with $c = a + b$.

Ted
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  • Okay, thank you for the quick answer. Just to be sure: Let $f_1, f_2 \in F$ be two elements of our basis, $f_1+f_2$ then is in $\mathbb{R}F$ as the sum of two elements of $F$ for which we do not define what the result of this sum actually is? – Ubuntix Nov 15 '20 at 22:04
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    The sum of $f_1$ and $f_2$ in $\mathbb{R} F$ is the formal sum $f_1 + f_2$ as an element of $\mathbb{R} F$, i.e., from your description of $\mathbb{R} F$, it is the element defined by $\lambda_{f_1} = 1$ and $\lambda_{f_2} = 1$ and $\lambda_x = 0$ for all $x \ne f_1, f_2$ (assuming $f_1 \ne f_2$). – Ted Nov 15 '20 at 22:06
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You're misunderstanding what the author denotes $\Bbb RF$. It is different from $F$. Actually,what he/she denotes this way (not a very good notation, in my opinion) is usually denoted as $$\Bbb R^{(F)},$$ which is the set of maps from $F$ into $\Bbb R$ with finite support. A well-known example is the vector space $\Bbb R^{(\Bbb N)}$, which is known, when endowed with the relevant multiplication (Cauchy product), as the ring of polynomials with real coefficients.

More generally, for any set $X$, one defines the free $K$-vector space with basis $X$ as $K^{(X)}$ (where $K$ is the base field). The construction of the author is further generalised to the situation where the base field is only a ring $R$ (in this case, it is called an $R$-module) and it can be used to prove that any $R$-module is the quotient of a free $R$-module (all modules, contrary to vector spaces, do not have a basis).

Bernard
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  • Could you clarify why the base field (in my case $\mathbb{R}$) is only a ring and not a field? – Ubuntix Nov 16 '20 at 15:37
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    I only meant that the generalisation of this construct is also valid for modules over a ring. I didn't say that $\mathbf R$ became a ring, but that the construct is valid in a more general situation. Afield is also a ring (+ some more properties). – Bernard Nov 16 '20 at 16:11
  • I had let your answer sink in for the last couple of days. I get the idea via the construction of our vector space as you have described.However, why can we write the set as all formal sum of elements of $F$ with real scalars? Shouldn't it then be the set of all formal sums of maps from $F$ into $\mathbb{R}$, e.g. if $f_1,f_2 \in \mathbb{R}^{(F)}$, then $f_1 + f_2 \in \mathbb{R}^{(F)}$, whereby $f_1, f_2$ are mappings? What's the reason for mapping into the real numbers, other than satisfying the vector space axioms? All sums would collaps into a single number.Would highly appreciate your help – Ubuntix Nov 18 '20 at 11:42
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    No they wouldn't collapse into a single number, for the same reason as, when you consider the vector space $\mathbf R^{3}$ and its canonical basis, $e_1+e_2=(1,1,0)$, not the number $2$. So to say, in $\mathbf R^{(F)}$, $e_f$ and $e_g$ do not live in the same copy of $\mathbf R$. – Bernard Nov 18 '20 at 13:36