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I am a mathematical illiterate, so I do not know what people mean when they say "equipped".

For example, I say that a Hilbert space is a vector space equipped with an inner product. What does that actually mean?

Obviously, one interpretation is to picture professor Hilbert as a plumber with an extra tool hanging out of his back pocket (a.k.a. an inner product), but mathematically why can't we do the inner product in a vector space?

Both Hilbert space and vector space work with functions and vectors, don't they?

Why can't we define a space where all operations are possible?

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    $\begingroup$ It just means "provided with". For example, a group is a set $G$ equipped with a function $G \times G \rightarrow G$ satisfying the group axioms. A topological space is a set $X$ equipped with a certain collection of distinguished subsets that satisfy the axioms for open sets in a topology. (By the way, your definition of a Hilbert space in the second sentence is, quite literally, incomplete. You forgot the condition that a Hilbert space is a vector space equipped with an inner product that makes the space complete.) $\endgroup$ – KCd Jan 18 '15 at 8:31
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    $\begingroup$ @KCd 'Provided with' has no mathematical meaning either. I've asked myself this in the past and if I was told what you say in your comment, I would have remained equally confused. $\endgroup$ – Git Gud Jan 18 '15 at 10:33
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    $\begingroup$ I enjoy the imagery of the inner product hanging out of a pair of pants. $\endgroup$ – Thoth19 Jan 19 '15 at 1:07
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    $\begingroup$ Your question is not clear. As I read it, the real question is why do we need to equip vector spaces with an inner product in the first place: "why can't we do inner product in vector space". The answer is simply that the axioms of vector spaces do not mention any inner product, and that (therefore) vector spaces do not in general carry an inner product. Some vector spaces cannot even be equipped with an inner product, and other could be so equipped, but in many differenet ways. $\endgroup$ – Marc van Leeuwen Jan 19 '15 at 5:26
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    $\begingroup$ Vector spaces over finite fields never have an inner product (because that requires a field of characteristic $0$). But if you want a real vector space that does not come equipped with an inner product, consider the space of rational functions in $X$. Or the formal power series in $X,Y$. Or the set of sequences satisfying some linear recurrence relation. Or for a smaller example the space of linear maps $\Bbb R^4\to\Bbb R^3$. Since you are asking something not to be given, there are just too many examples. $\endgroup$ – Marc van Leeuwen Jan 19 '15 at 21:52
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The word "equipped" keeps notational pandemonium from breaking loose. For instance, if you were to be a bit more formal, you'd say

A Hilbert space is a pair $(V, \left<\cdot,\cdot \right>)$, where $V$ is a vector space and $\left<\cdot,\cdot \right>\colon V\times V \to \mathbb{C}$ is an inner product. Additionally, all Cauchy sequences in $V$ are convergent in the norm induced by the inner product to an element in $V$.

But most of the time, there's no reason to disambiguate between the vector space and the inner product (who puts a different inner product on the set $L^2[0,1]$?), so we refrain from defining these "pairs", and simply "equip" our space.

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    $\begingroup$ For the sake of perspective: In my education we still use the tuples. We speak a group $(G,*)$ a ring $(R,+,.)$ a Boolean algebra $(B,\wedge,\vee,\neg)$, etc. $\endgroup$ – Syd Kerckhove Jan 18 '15 at 8:46
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    $\begingroup$ It is worth noting that your formal definition is not actually equivalent to "... a vector space equipped with a inner product." You also specify completeness. $\endgroup$ – Jørgen Fogh Jan 18 '15 at 13:45
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    $\begingroup$ The pandemonium really breaks loose when you note that a vector space is an abelian group equipped with an action by a field; and repeat for "abelian group" and "field". $\endgroup$ – Hagen von Eitzen Jan 18 '15 at 16:09
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Generally, when we use the word "equipped", we mean not that it is possible to do an operation, but we have one in mind. That is on a vector space like $\mathbb R^2$, we could think of various inner products that would suffice. For instance the following two operations are both inner products: $$(x_1,x_2)\cdot(y_1,y_2)=x_1y_1+x_2y_2$$ $$(x_1,x_2)*(y_1,y_2)=(x_1+x_2)(y_1+y_2)+x_2y_2$$ but they are different. That means that, when we choose an inner product, we are adding additional structure to $\mathbb R^2$ - just in a plain old vector space, we have no notion of orthogonal, and we can choose an appropriate inner product to make any pair of vectors orthogonal. For instance, in the above example, under the inner product $\cdot$, we have that $(0,1)$ and $(1,0)$ are orthogonal, but under the second inner product $*$ they have product $1$ and are not orthogonal (however, the vectors $(1,-1)$ and $(1,0)$ are, which is not true of the first metric). Similarly, notions like a unit vector cannot be defined in a vector space, as different inner products will give different answers. Thus, we must equip the vector space with some particular inner product before we can sensibly talk about an inner product - which means that we're keeping the old structure (of a vector space), but adding some new structure on top of it.

A notation, which is fairly prevalent in the field of general abstract nonsense, is to write something like "a set $G$ equipped with a binary operation $\cdot:G\times G\rightarrow G$" would be to refer to it as a tuple $(G,\cdot)$ - that is, we take a group (or magma) not to be just the set on which we have some external operation, but to take both set and operation into a single structure, as one is meaningless without the other.

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I think the best way of explaining what it means to be equipped with something is by showing how the term is used. Mathematics has a very precise formalism, but in its pure form it is mostly not appropriate for daily use because then even the simplest problems would become very hard. Equipping a mathematical object with additional structures is one of the tricks for allowing us to use shortcuts we are familiar with from daily life without giving up too much precision.

E.g., we could define a mounted policeman as a policeman P equipped with a horse H. Formally this means that a policeman is an ordered pair (P,H) consisting of a policeman and a horse. In reality there are some unstated assumptions that there is actually a certain connection between the policeman and the horse. (The policeman has to be sitting on it. Or at least he must have been assigned the horse for today's shift.) In mathematics, these tend to be more obvious and less ambiguous.

Technically we would have to distinguish strictly between the policeman P and the mounted policeman M = (P,H) of which he is the first constituent. E.g., if the policeman has a beard B(P), it just means that the first element of the mounted policeman (the policeman, not the horse or the pair) has a beard. Of course that's not how we say it in practice, and that's precisely the point. Even pure mathematicians will just write B(M) in most situations (rather than something like B(P(M))) because ultimately, we think of a mounted policeman as primarily a policeman. This becomes clear if you think of the tail of the mounted policeman. There is no such thing, even though the horse probably has one! This is why it's absolutely not the same to equip a policeman with a horse, or a horse with a policeman.

In fact, if we think of a policeman as a man who is equipped with a certain job, then saying that he has a beard really means that the man part m of the policeman has a beard. And if a man is a person equipped with masculinity and maturity, then for a man to have a beard really means that the person has a beard.

We can also think of a mounted policeman as a mounted policeperson who happens to be male, or as a mounted man who happens to have a certain job. Technically all these interpretations give rise to different definitions of a mounted policeman. But these differences are harmless and in most contexts (even most mathematical contexts) can be ignored: It is harmless to identify the different definitions of a mounted policeman.

In formulas, if m is the man, J is his job and H is his horse, then it doesn't matter in practice whether we think of the policeman as (m,J) and therefore of the mounted policeman as ((m,J),H), or if we think of the mounted man as (m,H) and therefore of the mounted policeman as ((m,H),J). So long as we remember where we have put the horse and where we have put the job, it's always clear how to translate from one representation to the other. Formally, a man equipped with a police job, and then equipped with a horse, is different from a man equipped with a horse and then equipped with a police job. But in practice we identify these two kinds of objects. I.e. we consider two mounted policemen to be the same if the underlying men are the same, the equipped horses are the same and the equipped jobs are the same.

The last point is different from natural language use, where we would consider M1 = (Chief Inspector Derrick, Rosie) and M2 = (Chief Inspector Derrick, Jack) to be the same mounted policeman. In mathematics we would instead be more precise and say that M1 and M2 are the same as (unequipped) policemen.

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    $\begingroup$ I think this is a wonderful answer, which comes closest to why mathematicians find it natural to speak the way we do. $\endgroup$ – Tom Church Jan 21 '15 at 21:35
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Well, sometimes we actually want to work with spaces were as few operations as possible are allowed. Often in math, we want to find the most general environment where a property holds, so we want to work in spaces with as few properties as possible. It's interesting to see just what we can do without needing to use an inner product, or a metric, or the lack of zero divisors to name a few. When you equip a vector space with an inner product, you are adding a property of the space that immediately makes the results less general. If you require an inner product, then the result can't be applied to general vector spaces.

If you defined a space where everything is allowed, then by definition it would not be a very interesting space.

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    $\begingroup$ Under my interpretation of the question, this answer misses the point completely. I too asked myself this in the past and I eventually learned that the answer (as per my interpretation of the question) is that 'set equipped with a certain apparatus' abbreviates a certain $n$-tuple. Perhaps the OP can clarify what he means with 'what does it actually mean?'... $\endgroup$ – Git Gud Jan 18 '15 at 10:29
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    $\begingroup$ The question seems to me to have (at least) two parts. It asks what "equipped" means. It also asks, "[W]hy can't we do inner product in vector space?" In other words, why wouldn't every vector space already have an inner product, making it redundant to "equip" it with one? This answer doesn't address the first part of the question, but it addresses the second part of the question, which most of the other answers (it seems to me) do not. $\endgroup$ – David K Jan 18 '15 at 15:16
  • $\begingroup$ @GitGud I was aiming more to explain, like David said, why we even bother to define spaces with fewer allowed operations. I'm well aware that this does not explain what "equipped" means :) $\endgroup$ – Johanna Jan 18 '15 at 15:58
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Fundamentally, "equipped with" just means "and". A Hilbert space is an object which consists of two components: a vector space, and an inner product on that vector space.

This means that strictly speaking, a Hilbert space and a vector space are different objects: a Hilbert space is not a vector space, and a vector space is not a Hilbert space.

But the reason that a mathematician writes "equipped with" is that they don't want to speak strictly; they want to speak loosely, because that's easier. So what your sentence means is this:

A Hilbert space is an object which consists of two components: a vector space, and an inner product on that vector space. However, I expect it to be convenient to speak as if a given Hilbert space were the same thing as the vector space which is its main component, and so I will do so.

And so, thenceforth, whenever the mathematician talks about a Hilbert space $V$, they will often speak as if $V$ were a vector space, rather than merely having a vector space as a component.

As for why you can't do inner products in vector spaces in general? Well, nobody has ever defined the word "inner product" in a way which makes sense for every vector space. You can't take an inner product until you've decided what the phrase "inner product" means in your specific case.

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    $\begingroup$ Which is to say that 'equipped' is metaphorical language, not a technical term. $\endgroup$ – Mitch Jan 19 '15 at 20:44
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Informally, let's consider a mathematical space, which we think of primarily as a set $A$ of points (or primitive elements) on which are defined various operations (e.g. addition, scalar multiplication with a field, scalar product, etc.). We can say that $A$ is "equipped" with these operations. That is, whenever we mention $A$, we have in mind these operations (and not other operations that might well be performed on $A$ in other circumstances).

This idea of "equipping" $A$ with operations is very easy to understand intuitively and use in practice. But, to be sure that this idea is rigorously correct, we need to formalize it in set theory. The simplest way to do this is to define the space "equipped with" (sometimes "together with") a structure as a tuple: $(A,f,g,h,...)$, where $f,g,h,...$ are the relevant operations or other sorts of structure. So far, easy enough. The problem starts when we try to access the thing we are really wanting to discuss: the set $A$; we are not interested in talking about tuples. It is straightforward to assemble $(A,f,g,h,...)$, starting with $A$, by using a chain of Kuratowski pairings. But to get from $(A,f,g,h,...)$ to $A$, set-theoretically, requires a reverse chain of examining two-element sets, determining which of the two elements has only one element, and picking out that element as the starting set for the next stage of the chain.

We can cut through most of this set-theoretic labour by quoting a general theorem of set theory that "the first component" of a tuple is well defined. But why even bother to do that? it's much nicer to drop the whole formalistic business, including the "$(A,f,g,h,...)$" notation, and use our informal idea of "equipped", knowing that we could in principle formalize it if we had to do so.

Authors may refer to both $A$ and $(A,f,g,h,...)$ by the same word, e.g. "space". This is unfortunate; but a certain amount of looseness in mathematical description has to be forgiven to avoid bogging the text down with repetitive distracting pedantic detail that the experienced reader doesn't need.

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