Is it generally implicitly assumed that if $B$ is a subalgebra of a unital Banach algebra $A$ then $1 \in B$?

I tried to find a definition of subalgebra but the only definition I found was in Murphy who defines it to be ''itself a normed algebra with the norm got by restriction''

It seems to be that here at least $1 \in B$ is not required. It's also easy to come up with examples: Let $A = \mathbb C^2$ and $B=$ one the axes (unit $=(1,1)\notin B$).

Nevertheless, I have doubts. Are there situations where it is desirable to consider non-unital subalgebras of unital algebras?


First, an answer to your question. No, one typically does not assume that Banach subalgebras are unital. One of the standard examples of a Banach algebra is $C(X)$ where $X$ is compact, and one of the standard examples of a Banach subalgebra of $C(X)$ (an ideal, in fact) is is $C_0(U)$ where $U \subseteq X$ is an open set. But $C_0(U)$ is not unital.

Second, a comment. In Banach algebra theory you do not need to worry about units the way you do in most of ring theory simply because unitalizations behave very nicely. If $A$ is a unital Banach algebra and $B$ is a non-unital Banach subalgebra the unitalization $\tilde{B}$ sits naturally in $A$ (not true for arbitrary rings, even commutative ones).

  • $\begingroup$ Thinking about it now, my comment about unitalizations might only be true for C*-algebras. I'm not completely sure. $\endgroup$ – Paul Siegel Mar 25 '14 at 13:19
  • $\begingroup$ Thank you for your answer. I'm a bit confused now: how can a function that is only defined on a subset $U$ of $X$ be an element of $C(X)$? $\endgroup$ – Student Mar 26 '14 at 7:46
  • $\begingroup$ @Student The notation $C_0(U)$ means that every $f \in C_0(U)$ vanishes at infinity; thus the function $g$ on $X$ defined by $g(x) = f(x)$ for $x \in U$ and $g(x) = 0$ for $x \in X - U$ is continuous. This is how $C_0(U)$ embeds in $C(X)$. $\endgroup$ – Paul Siegel Mar 26 '14 at 19:08

Ultimately, it is just a matter of convention and how a particular author wants to write things. Authors usually choose the path that makes their writing more succinct and easier to read.

The example you gave is a common one: Make a product of objects... shouldn't the original objects be subobjects of the new object? Depending on your decision, the definition will be one or the other. For another example different from just ideals of rings, consider any big von Neumann ring $R$. The thing is filled with idempotents, and for each nontrivial idempotent $e\in R$, $eRe$ is again a ring (with identity $e\neq 1$!)

But let's also keep in mind that there are a great many arguments that do depend on sharing the identity, and choosing that convention is the best for them.

I think people often get too stressed about this thing with identities in particular. Just remember that definitions are (most often) not a matter of correct or incorrect but more of a matter of clarity and ease of exposition. One definition or notation is "better" than another if it is relatively easily understood and minimizes confusion. The mistake many people make is thinking that there is always a universally "best" definition/notation.

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    $\begingroup$ Definitions are usually chosen with a specific set of examples in mind. Because of this it is hard to imagine a good reason to insist that Banach subalgebras have units. $\endgroup$ – Paul Siegel Mar 25 '14 at 13:32

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