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Let $(\mathcal A,||\cdot||)$ be a normed algebra (with or without a unit). The unitization of $\mathcal A$ is the space $\mathcal A_+:=\mathcal A\oplus \Bbb C$ where the multiplication operation $\cdot$ and norm $|||\cdot |||$ are defined by $$\begin{align} (a,\lambda)\cdot(b,\mu) &:=(ab+\mu a+\lambda b,\lambda\mu) \\ |||(a,\lambda)||| &:= ||a||+|\lambda| \end{align}$$ for any $a,b\in\mathcal A$ and $\mu,\lambda\in \Bbb C$. It can be easily verified that $(0,1)$ is a unit in $\mathcal A_+$ with norm $1$, and that $\mathcal A$ isometrically embeds into $\mathcal A_+$. This construction seems to be a very fundamental tool used to augment a unit into a space without one , it came up in my first class of the chapter on Banach Algebra.

Our professor tried to convince us that unitization is done in the same spirit as completion of metric spaces. However, this construction troubles me for 2 main reasons.

Firstly, nothing stop me from unitizing a unital normed algebra. This gives me an essentially different space than the one I started with. This is not the case for the completion $\hat X$ of a Banach space $X$, in which we have $\hat X \cong X$. To add salt to the wound, the original unit in $\mathcal A$ is no longer a unit in $\mathcal A_+$.

Secondly, unlike the process of completion, the adjoined unit seems quite artificial. In metric space completion we only fill in "holes" but it seems like in unitization we artificially add genuine new direction into our space.

I know that my professor's analogy shouldn't be taken literally but in what sense does unitization resemble metric completion? Conceptually, what is unitization?

It also occurred to be that I'll learn to appreciate unitization as I encounter more and more results in this field. Nevertheless, I would like to have an intuitive understanding of unitization.

For example, take $(\mathcal A,+,\cdot,||\cdot||)$ to be $(L^1(\Bbb R),+,*,||\cdot||_{1})$, where $*$ is the convolution. This is a non-unital algebra since the Dirac distribution does not belong to $L^1$. In this case what does $\mathcal A_+$ look like?

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    $\begingroup$ Just to address the very last question: the unitization is $L^1({\bf R})+{\bf C}\delta_0$ sitting inside $M({\bf R})$. $\endgroup$
    – Yemon Choi
    Commented Nov 17, 2016 at 22:54
  • $\begingroup$ Briefly: one can describe this "making-a-unit" process in a somewhat better way which does indeed have the feature that if the thing already has a unit, then nothing happens. After some decades of confusion about this in my own mind, my conclusion is that one should describe the process in naive-categorical terms (in a sense analogous to naive set theory, as opposed to axiomatic), specifically to say what the intent is, that is, what we want/need of the "unitization". Just "making things" is not necessarily purposeful or intelligible, indeed, as is (to my mind) implicit in this question. $\endgroup$
    – paul garrett
    Commented Nov 17, 2016 at 23:10
  • $\begingroup$ @paulgarrett I am not sure if I understand you correctly, but did you mean that there is some point of view in which the new unit coincide with the original one (if it exists)? If that's the case could you please elaborate a little bit more? $\endgroup$
    – BigbearZzz
    Commented Nov 17, 2016 at 23:22
  • $\begingroup$ Getting late for me... "an older person"... but, yes, there is a way to characterize "unitization" (or whatever word you like) so that it does nothing to an already-having-a-unit algebra. Maybe Rudin's "Fun. An." does this, although I have a little bit of apprehension about that source, namely, that maybe is was not so forthright about the function of things... $\endgroup$
    – paul garrett
    Commented Nov 17, 2016 at 23:35
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    $\begingroup$ @BigbearZzz: for each $a \in \mathcal{A}$ define $M_a: \mathcal{A} \to \mathcal{A}$ by $M_a(b) = ab$. Then $M_a \in B(\mathcal{A})$ and $\|M_a\| \leq \|a\|$; if $\mathcal{A}$ is regular then $\|M_a\| = \|a\|$ for all $a$. In this case you can define $\tilde{\mathcal{A}}$ to be the span of $\{M_a: a \in \mathcal{A}\}$ and the identity operator. If $\mathcal{A}$ has a unit then you haven't added anything. $\endgroup$
    – Nik Weaver
    Commented Nov 18, 2016 at 15:34

3 Answers 3

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Unitization and metric completion are both left adjoint functors, as are may other "-tion" operations in mathematics, such as localization or abelianization. Specifically, there is a forgetful functor from unital algebras to nonunital algebras (including norms is not particularly important here), and unitization is its left adjoint. Conceptually this means that it is in some sense the "freest" or "laziest" way of adding a unit to a nonunital algebra. (It's important to keep in mind that "nonunital" means "not necessarily having a unit" here.)

An important difference between unitization and metric completion is that the inclusion of complete metric spaces into all metric spaces is fully faithful, but the inclusion of unital algebras into nonunital algebras is not. This is responsible for your observation that when you complete a complete thing nothing happens, but when you unitize a unital thing you get something different. One of the relevant keywords here is idempotent monad.

One way of thinking geometrically about what unitization accomplishes is to apply the commutative Gelfand-Naimark theorem to nonunital C*-algebras. This says that every nonunital C*-algebra is the algebra of functions vanishing at infinity on some locally compact Hausdorff space $X$. Taking the unitization then gives the algebra of functions on the one-point compactification $\hat{X}$ of $X$. Taking the unitization again gives the algebra of functions on the one-point compactification of $\hat{X}$, which is $\hat{X}$ with a disjoint point added.

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  • $\begingroup$ Incidentally, in order to make metric completion an actual left adjoint the morphisms between metric spaces need to be weak contractions, also known as short maps; those maps $f$ satisfying $d(f(x), f(y)) \le d(x, y)$. See ncatlab.org/nlab/show/metric+space#lawvere_metric_spaces for some discussion of this choice. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 18, 2016 at 1:42
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    $\begingroup$ Nice answer. I think "one-point compactification" is a much better analogy for unitization than "completion". $\endgroup$
    – Nate Eldredge
    Commented Nov 18, 2016 at 2:29
  • $\begingroup$ "Specifically, there is a forgetful functor from unital algebras to nonunital algebras(...)" -- I think it would be more accurate to say that the functor is to all algebras. If I'm wrong, I would like to know what functor you had in mind. :) $\endgroup$
    – tomasz
    Commented Nov 18, 2016 at 5:30
  • $\begingroup$ Probably one should explicitly mention the fact that adding a unit might be useful when you study maps between unital algebras that do not preserve the unit: you can extend these to unital maps between unitizations. $\endgroup$
    – Mateusz Wasilewski
    Commented Nov 18, 2016 at 6:10
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    $\begingroup$ @tomasz: I use "nonunital" to mean "not-necessarily-unital," compatibly with the convention where "noncommutative" means "not-necessarily-commutative," "nonassociative" means "not-necessarily-associative," etc. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 18, 2016 at 7:58
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I don't remember where I read this, but Gert Pedersen once said something to the effect that "When I was young, the first thing we did with any C*-algebra was to adjoin a unit, but nowadays the first thing we do is remove the unit by tensoring with the compacts." The point is that $\mathcal{A}\otimes \mathcal{K}$ is the "stabilization" of $\mathcal{A}$: it is isomorphic to the $n\times n$ matrices over itself, for any $n$.

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  • $\begingroup$ Sadly I can only relate to the first half of your quote. :) I'd really love to know what "tensoring with the compacts" is, though. $\endgroup$
    – BigbearZzz
    Commented Nov 18, 2016 at 15:24
  • $\begingroup$ Not sure what your background is $\ldots$ let $\mathcal{A}\otimes \mathcal{K}_0$ be the set of all infinite matrices, with all entries in $\mathcal{A}$ and only finitely many nonzero entries. If $\mathcal{A}$ acts on $H$ then $\mathcal{A}\otimes \mathcal{K}_0$ acts on $H\otimes l^2$. $\mathcal{A}\otimes \mathcal{K}$ is the completion of this object. $\endgroup$
    – Nik Weaver
    Commented Nov 18, 2016 at 15:37
  • $\begingroup$ FWIW: that "quote" rang a bell, and some searching shows it's in the introduction of Olsen--Pedersen, Corona C*-algebras and their applications to lifting problems, Math. Scand. 64 (1989), 63--86. $\endgroup$
    – Yemon Choi
    Commented Dec 8, 2016 at 21:42
  • $\begingroup$ Cool. Wow, it's been decades since I looked at that paper. $\endgroup$
    – Nik Weaver
    Commented Dec 9, 2016 at 1:14
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When $A$ is a C*-algebra, the norm $$\Vert (a,\lambda)\Vert := \Vert a\Vert + |\lambda|$$ does not make $A_+$ a C*-algebra because the axiom $\Vert a^*a\Vert = \Vert a \Vert^2$ will no longer hold. So one usually replaces this norm by $$\Vert (a,\lambda)\Vert := \sup_{b\in A,\ \Vert b\Vert \leq 1}\Vert ab+\lambda b\Vert.$$
In case $A$ is already unital, with unit $1_A$, the above definition will give $\Vert (1_A,-1)\Vert = 0$, so this process only works when $A$ is not unital, meaning neccessarily without a unit.

One may thus interpret this as saying that unital C*-algebras do not like to be unitized :-)

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  • $\begingroup$ However, a norm which works both in the unital and non-unital case may be found in Johannes Hahn's answer to this question $\endgroup$
    – Ruy
    Commented Nov 19, 2016 at 14:44
  • $\begingroup$ I think this is more or less what I explained in my comment to the original question (since every C*-algebra is regular). $\endgroup$
    – Nik Weaver
    Commented Nov 19, 2016 at 15:20
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    $\begingroup$ Another comment, it's my understanding that K-theory people actually like to add units to C*-algebras that are already unital, so that the transformation $\mathcal{A} \mapsto \mathcal{A}^+$ is functorial. (It's easy enough to make $\mathcal{A}^+$ a C*-algebra: put $\mathcal{A}$ in $B(H)$, embed in $B(H\oplus \mathbb{C})$, and add the identity operator of the latter.) $\endgroup$
    – Nik Weaver
    Commented Nov 20, 2016 at 0:30

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