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Determine all the maximal ideals and all the characters in $\ell^1(N)$, if we know that it is a commutative Banach algebra with component-wise multiplication and addition.

I know what are characters in $\ell^1(Z)$ but don't know how to determine them in $\ell^1(N)$. Any help would be appreciated.

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  • $\begingroup$ are you sure that the product you want to consider is not the convolution product instead of component-wise multiplication? If with $N$ you mean $\mathbb{N}$, then $l^1(\mathbb{N})$ with pointwise multiplication is not an algebra (it is not closed under this multiplication), but it is if you consider the convolution as a product. $\endgroup$ – Bob Jul 25 '18 at 5:36
  • $\begingroup$ @Bob Contrary to your claim, $\ell^1(\mathbb N)$ is closed with regards to pointwise multiplication. For $x,y\in \ell^1(\mathbb N)$, we have $\lVert x\rVert=\sum_{n\in\mathbb N}\lvert x_n \rvert<\infty$ and similarly $y$; $\lVert xy \rVert=\sum_{n\in\mathbb N} \lvert x_n\rvert \lvert y_n\rvert\leq \sum_{n\in\mathbb N}\lvert x_n \rvert \sum_{n\in\mathbb N}\lvert y_n \rvert<\infty$. $\endgroup$ – suhogrozdje Aug 5 '18 at 15:28
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    $\begingroup$ @suhogrozdje : probably I was in alcoholic coma and i got confused with $L^1$... thx for having pointed out :) $\endgroup$ – Bob Aug 5 '18 at 15:40
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Short answer: Even though $\ell^1$ is a non-unital Banach algebra, maximal ideals are precisely the kernels of characters. The nontrivial characters $\{\varphi_N\in\hat {\ell^1}\,;\,N\in\mathbb N\}$ are given by $\varphi_N(a_i)_i=a_N$.

Proof.

  1. First note that $\ell^1$ is a non-unital commutative Banach algebra, because the norm of the unit $(1)_{i\in\mathbb N}$ is infinite. Then maximal ideals and characters are not necessarily in a bijective correspondence. However, for an arbitrary non-unital commutative Banach algebra $A$, it can be shown that the map $\phi\colon \hat A\rightarrow \mathcal M(A)$, $\phi(\varphi)=\ker\varphi$ is injective (here, $\hat A$ denotes the set of all characters of $A$ and $\mathcal M(A)$ the set of all maximal ideals in $A$). To see this, let $\varphi,\psi\in\hat A$ be nontrivial with coinciding kernels and take $a\in A\setminus\ker\varphi$. We need to show $\varphi\equiv\psi$. For any $b\in A$, we have $$b-\frac{\varphi(b)}{\varphi(a)}a\in\ker\varphi=\ker \psi$$ and so $\psi(b)=\frac{\varphi(b)}{\varphi(a)}\psi(a)$; plugging $b=a^2$ gives us $\varphi(a)=\psi(a)$, which is enough to show, because $A=\ker\varphi + \mathrm{Lin}(a)$. The latter holds because any $b\in A$ can be written as $b=\frac{\varphi(b)}{\varphi(a)}a+\left(b-\frac{\varphi(b)}{\varphi(a)}a\right)$. It also follows from this that $\ker\varphi$ is a maximal ideal (what happens if an ideal containing $\ker\varphi$ contains a nonzero element of $\mathrm{Lin}(a)$?).
  2. Now let's find all characters of $\ell^1$. For a character $\varphi\in\hat{\ell^1}$, we have by multiplicativity $\varphi(ab)=\varphi(a)\varphi(b)$ for any $a=(a_i)_i, b=(b_i)_i\in\ell^1$. If we take $a=e^N$, $b=e^M$ for $M\neq N\in\mathbb N$ (here $e^N=(\delta_{i,N})_i$, i.e. the canonical basis vector), we get $$ 0=\varphi(e^N)\varphi(e^M)\implies \varphi(e^N)=0\text{ or } \varphi(e^M)=0. $$ Also, taking $a=b=e^N$, we get $\varphi(e^N)=1$ if $\varphi(e^N)\neq 0$. It follows from this that any nontrivial character sends exactly one basis vector to 1 and all other basis vectors to 0, i.e. $$ \hat {\ell^1}=\{\varphi_N\,;\,N\in\mathbb N\}\cup\{c_0\} $$ where $\varphi_N(e^N)=1$ and $\varphi_N(e^M)=0$ for all $M\neq N$ and $c_0$ is the trivial character. Note that $\varphi_N$ is uniquely determined by where it maps the basis vectors, which is due to its linearity. $\varphi_N$ is bounded, since $\varphi_N((a_i)_i)=a_N\leq \lVert a \rVert$ and its kernel is obviously $$ \ker\varphi_N=\{(a_i)_i\in\ell^1\,;\,a_N=0\}. $$
  3. By the first point, we have that $\{\ker\varphi_N\,;\,N\in\mathbb N\}\subset \mathcal M(\ell^1)$. Does the inclusion reverse? Let $I\in\mathcal M(\ell^1)$ such that $I\not\subset \ker\varphi_N$ for all $N\in\mathbb N$. Then $$ \forall N\in\mathbb N\colon\exists u^N=(u_i^N)_{i}\in I\colon u_N^N\neq 0 $$ and because $I\triangleleft \ell^1$, we have $\forall (a_i)_{i}\in\ell^1\colon (a_iu_i^N)_{i}\in I$. Taking $a_N=(u_N^N)^{-1}$ and $a_i=0$ for all $i\neq N$, we see that $e^N\in I$ for all $N\in\mathbb N$ and because $I$ is a linear subspace of $\ell^1$, this implies $I=\ell^1$. Thus the inclusion reverses.
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  • $\begingroup$ @XYZ Is there anything you'd like me to add to make this an acceptable answer? $\endgroup$ – suhogrozdje Aug 13 '18 at 13:55

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