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On Singer-Wermer conjecture in Frechet algebras

We construct two Frechet algebras admitting countably many mutually inequivalent Frechet algebra topologies. The second example is a modification of the maiden (and first) example of a non-Banach Frechet algebra with two inequivalent Frechet algebra topologies, considered by Read to show that the famous Singer-Wermer conjecture does not hold in the Frechet case. However the conjecture does hold by the second example which also admits countably many mutually equivalent Frechet algebra topologies.


Introduction
Recently, the author posed a question about the existence of a Fréchet algebra with infinitely many inequivalent Fréchet algebra topologies [8,Question 4].
This question carries a lot of importance due to the following three reasons.
First, in automatic continuity theory, we would normally like to examine when and how the algebraic structure of the algebra A determines the topological structure of A, in particular, the continuity aspect (and more particularly, the uniqueness of the topology of A; see [2,7,8,9] for more details). So it is natural to expect that the non-uniqueness of the topology would reflect some properties of the algebraic structure of A.
In [4], Feldman constructed an example to show the failure of the Wedderburn Theorem in the Banach algebra case. This algebra is ℓ 2 ⊕ I C as a vector space with the usual product in ℓ 2 and trivial multiplication by the second summand, and has a norm in which ℓ 2 is dense. In [10], Read constructed an example (see §2 below) to show the failure of the Singer-Wermer conjecture (the commutative case) in the Fréchet algebra case; this conjecture holds in the Banach algebra case (see [11]). Thus the situation on Fréchet algebras is markedly different from that on Banach algebras, and that a structure theory for Fréchet algebras behaves in a very distinctive manner from Banach algebras; for example, the prestigious Michael problem is still remained unsolved since 1952.
Second, it is easy to give uncountably many inequivalent Fréchet space topologies to a very familiar Fréchet space, namely, spaces of holomorphic functions (see [12] for more details). However the same space is also a semisimple Fréchet algebra of power series, and so, it admits a unique Fréchet algebra topology (see [8] for more details). Thus it is expected that the case of having infinitely many inequivalent Fréchet algebra topologies by some Fréchet algebra should be rare, and the former situation should be common.
We note that in [6], Loy gave a method for constructing commutative Banach algebras with inequivalent complete norm topology. We exploit the idea of using the discontinuity of derivations to give a Fréchet algebra other inequivalent Fréchet algebra topologies. In view of a comment in the previous paragraph, it is interesting to recall an early exposition of Banach algebras of powers series and of their automorphisms and derivations, given by Grabiner in [5]. We remark that although the Fréchet algebra topology τ 0 + τ 0 (defined below) of F ∞ ⊕ F ∞ is not obtainable by our approach, other inequivalent Fréchet algebra topologies on F ∞ ⊕ F ∞ may be constructed.
The purpose of this paper is to show that commutative Fréchet algebras F ∞ and F ∞ ⊕ F ∞ admit countably many mutually inequivalent Fréchet algebra topologies.
Third, we do not know any non-trivial examples of (commutative) Banach algebras with infinitely many inequivalent complete algebra norms till this date (see Conclusions below). Of course, one may consider the trivial product on a Banach space with countably many inequivalent complete algebra norms to have such an example in the Banach algebra case.

First Example
In [10], Read showed that the algebra F ∞ = I C[[X 0 , X 1 , . . .]] of all formal power series in infinitely many commuting indeterminates X 0 , X 1 , . . . has two inequivalent Fréchet algebra topologies τ 0 and τ c . The natural derivation ∂/∂X 0 is continuous with respect to the Fréchet algebra topology τ c . He also showed that the natural derivation ∂/∂X 0 is discontinuous with respect to the Fréchet algebra topology τ 0 and with the image the whole algebra, and that X 0 lies in the closure of the coefficient algebra In fact, it is surprising to observe that for each i ∈ I N , the other natural derivation ∂/∂X i is discontinuous with respect to the Fréchet algebra topol-ogy τ i (defined below), and that for each i ∈ I N , X i lies in the closure of the such that X i does not belong to A i . Indeed, with respect to τ i one has X n → X i as n → ∞; so one has Thus 1 is in the separating subspace for ∂/∂X i ; since the subspace is an ideal, it is the whole algebra. This shows that for each i ∈ I N , ∂/∂X i vanishes on a dense subset A i of (F ∞ , τ i ). Not only this, but it is easy to see that for i = j, the Fréchet algebra topologies τ i and τ j on F ∞ are not mutually comparable, because the identity map is con- Hence F ∞ admits countably many mutually inequivalent Fréchet algebra topologies. Moreover, for i = j, the natural derivations ∂/∂X i are continuous on (F ∞ , τ j ) as X n → X j and ∂/∂X i (X j − X n ) = 0.
We shall feel free to use the terminology and conventions established there in [10]. Not only this, but our argument for this section is kept short because it uses the key ideas involved in defining the Fréchet algebra topology τ 0 on F ∞ . Let us therefore define ourselves bizarre topology τ i on F ∞ for each i ∈ I N . For this, first, let us choose a linear functional ψ : A (1) → I C with the property that ψ(X 0 ) = 0 and ψ = X * n on A (1, 0) = B (1, 0) (the subspace of A (1) , consisting of all formal sums ∞ j = 1 λ j X j ), i.e., the coordinate functional such that X * n ( ∞ j = 1 λ j X j ) = λ n (and hence ψ(X n ) = 1 for all n ∈ I N ). Note that this requires extending the sequence (X n ), n ∈ I N 0 , to a Hamel basis of A (1) , and thus involves the axiom of choice. Then let (φ n ), n ∈ I N 0 , be linear functionals on A (1) as follows: (a) For n ∈ I N 0 , n = i > 0, φ i = ψ, the "discontinuous" linear functional, defined above. (b) For n ∈ I N 0 , n = i, φ n = X * n , the coordinate functional such that X * n ( ∞ j = 0 λ j X j ) = λ n . For each i > 0, let τ i be the locally multiplicative convex topology on B (a non- )| for all a ∈ B, for this sequence of linear functionals φ n (for i = j, the Fréchet algebra topologies τ i and τ j are mutually inequivalent, as described above). We remark that our discontinuous linear functional ψ here is different from Read's discontinuous linear functional (see [10,Definition 1.7]). In fact, the modification in the definition of ψ is needed so as to change the position of φ i = ψ in the sequence (φ n ) of linear functionals, in order to ultimately generate mutually inequivalent Fréchet algebra topologies τ i on F ∞ .
We claim that B is, in fact, complete under the topology τ i for each i > 0, so (B, τ i ) is a Fréchet algebra. Since the averaging map α : B → B, the locally finite map such that α( is a τ i -continuous projection, the subspace A = α(B) = ker(I −α) is closed, so (A, τ i ) is also a Fréchet algebra. Note that if we prove our claim we have our for all n, m with n < N. Hence, X N → X i in τ i . One may then use the derivation ∂/∂X i as above, and the separating subspace is the whole algebra, be it F ∞ or B. All that remains is to prove B complete.
First, we note that (B, τ c ) is a Fréchet algebra; to define τ c , apply the method of Corollary 1.11 of [10] to the sequence of coordinate function- To prove B complete, by using the seminorms · (m) n , defined above, we show that the linear map Ψ : B → B, defined by . . , i r ) ∈ I N r 0 such that exactly k of the i j are equal to zero and let B (r,k) be the subspace consisting of all power series of the form i∈F (r,k) a i X ⊗i . Then B (r) = ⊕ r k=0 B (r, k) . Let β r, k be the projection onto B (r, k) parallel to the others. We claim that for each k, β r, k Ψ (r) | B (r, k) is equal to the identity map on B (r, k) ; and we claim that Ψ (r) maps B (r, k) into ⊕ r l=k B (r, l) . Thus we shall show that the action of Ψ (r) is "lower triangular" and nonsingular, with respect to the decomposition . Therefore, Ψ (r) -and hence Ψ -is bijective.
In view of definition of Ψ above, β r, k Ψ (r) | B (r, k) is equal to the identity map if and only if we have for each l = (l 1 , . . . , l r ) ∈ F (r, k) and a ∈ B (r, k) ; and Ψ (r) maps B (r, k) into for all m < k, l = (l 1 , . . . , l r ) ∈ F (r, m) and a ∈ B (r, k) . Let us prove (1) and (2) together, by induction on r.
When r > 1, we proceed by induction. First we establish statements (1) and (2) when r = 2. In this case, from Definition 1.9 of [10], we have the "tensor product by rows" To prove statement (1), consider the case when l = (l 1 , l 2 ) ∈ F (2, k) and a ∈ B (2, k) . Note that P j a lies in the division on the left by X 0 removes a factor X 0 from each monomial X ⊗i .
If on the other hand, l 1 > 0, then we have two cases: In this case, l 2 ∈ F (1, 1) and P i a ∈ B (1,1) . So, the statement (3) is equal to In this case, (l 2 ) ∈ F (1,k) and P l 1 a ∈ B (1,k) so the statement (3) is equal to (P l 1 a) l 2 = a l 1 , l 2 as before.
For this, follow Read's argument in our case.
and its separating subspace is all of B. The derivation ∂/∂X i : (A, τ i ) → (A, τ i ) is also discontinuous, and its separating subspace is all of A.
Proof. Follow the proof of [10,Theorem 2.5] in this case and the result follows. ✷

Second Example
We now show that the algebra F ∞ ⊕ F ∞ has countably many mutually inequivalent Fréchet algebra topologies. In fact, the algebra F ∞ has been completed by the adjunction of a radical so that F ∞ ⊕ F ∞ has Fréchet algebra topologies in which F ∞ is dense in those topologies.
Let F ∞ be a commutative Fréchet algebra with Fréchet algebra topology τ i generated by the sequence (p ′ k, i ) for each i ∈ I Z + , by §2 and τ c generated by the sequence (p k ). Let A denote the vector space direct sum F ∞ ⊕ F ∞ with product (x, m)(y, n) = (xy, x · n + y · m) and seminorms generating the Fréchet algebra topology τ c + τ c , and seminorms generating the Fréchet algebra topology τ i + τ i for each i ∈ I Z + . For each is defined on the algebra F ∞ ⊕ F ∞ and is, in fact, easily seen to be a submultiplicative seminorm thereon; for each i ∈ I Z + , the sequence (q k,∂ i ) generates the Fréchet algebra topology τ ∂ i equivalent to the Fréchet algebra is defined on the algebra F ∞ ⊕F ∞ and is, in fact, easily seen to be a submultiplicative seminorm thereon; for each i ∈ I Z + , the sequence (q ′ k,∂ i ) generates the Fréchet algebra topology τ ′ ∂ i inequivalent to the Fréchet algebra topology ) and so extends uniquely to a map of the completion ( ) is a seminorm on F ∞ and θ ∂ i • ι extends to an isometric isomorphism of (F ∞ ) ∂ i , the completion of F ∞ under (q ′′ k,∂ i ) (respectively, (q ′′′ k,∂ i )), with Gr(∂ i ), the closure (respectively, in (F ∞ ⊕ F ∞ , τ c + τ c ) and in (F ∞ ⊕ F ∞ , τ i + τ i )) of the graph of ∂ i .
Let i ∈ I Z + be fixed. Now if ∂ i is continuous, then (q k,∂ i ) is equivalent to This is the case when F ∞ is a Fréchet algebra under the Fréchet algebra topology τ c . In the discontinuous case q ′′′ k,∂ i is a discontinuous seminorm on (F ∞ , τ i ) and ι is a discontinuous isomorphism. We study this case in detail.
First, by Theorem 3.2 below, (q Suppose now that ∂ i is discontinuous. This is the case when F ∞ is a Fréchet algebra under the Fréchet algebra topology τ i , by §2. Then we have F ∞ has a completion with a non-trivial nil ideal, that is, the completion ing theorem. We remark that if A is a semisimple Fréchet algebra, then [4] for details), and in particular, if A is a pro- and which is inequivalent to the Fréchet algebra topology τ i + τ i , generated by (q ′ k, i ). In particular, the algebra F ∞ ⊕ F ∞ admits countably many Fréchet algebra topologies inequivalent to the Fréchet algebra topology τ i + τ i .
Proof. First, by §2, we note that, for each i ∈ I Z + , ∂ i vanishes on a dense subset A i (which is the coefficient algebra of F ∞ ); we also note that this is possible due to the unique property of the Fréchet algebra topology τ i as explained in §2. Now follow the proof of [6, Theorem 1] for this case. ✷ B. Theorem 3.1 gives countably many mutually inequivalent Fréchet algebra topologies on F ∞ ⊕ F ∞ ; for mutual inequivalence, see Theorem 3.5 below.
In fact, we have a more stronger result than Theorem 3.1 as follows. We note that, for each i ∈ I Z + , the derivation ∂ i on F ∞ induces a natural derivation if and only if for each i ∈ I Z + , the natural derivation D i is continuous on ) is a Fréchet algebra. Let ((a n , x n )) be a Cauchy sequence in (F ∞ ⊕ F ∞ , (q k,∂ i )) (respectively, for each i ∈ I Z + , ). Then (a n ) and (∂ i (a n ) − x n ) are Cauchy sequences in (F ∞ , (p k )) (respectively, for each i ∈ I Z + , (F ∞ , (p ′ k, i )). Since F ∞ is a Fréchet space, there exists a ∈ F ∞ and x ∈ F ∞ such that a n → a and ∂ i (a n ) − x n → x. Then (a n ,

2.
It is easy to see that for each i ∈ I Z + , the Fréchet algebra topologies τ c + τ c and τ i + τ i are inequivalent, since if they are equivalent on the algebra F ∞ ⊕ F ∞ , then since F ∞ can be identified with the closed subalgebra Proof. Clearly, by Theorem 3.2, for each i ∈ I Z + , the Fréchet algebra topology τ ∂ i generated by the sequence (q k,∂ i ) is equivalent to the Fréchet algebra topology τ c + τ c , since the derivations ∂ i is continuous on (F ∞ , τ c ). Now, the Fréchet algebra topology τ ∂ i is equivalent to τ ∂ j by the transitivity property of the relation "equivalence of topologies". ✷ In fact, we can sharpen the above corollary as follows.
Corollary 3.4 Let i and j be in I Z + such that i = j. The Fréchet algebra topology τ ∂ i is equivalent to the Fréchet algebra topology τ ∂ j on the algebra Proof. Suppose that for i = j, the natural derivation D i is continuous on  , x)). Hence since the sequences (q k,∂ i ) and (q k ) are equivalent and since D j is continuous on (F ∞ ⊕ F ∞ , τ c + τ c ). Thus, for i = j, the natural derivation D i is continuous on (F ∞ ⊕ F ∞ , τ ∂ j ) and the natural derivation Proof. Let (F ∞ ⊕ F ∞ , τ ′ ∂ j ) be a commutative Fréchet algebra and let D i be the discontinuous natural derivation on F ∞ ⊕ F ∞ , induced by ∂ i . Then, following the arguments of Theorem 1 of [5], . Now a diagonal argument shows that S is closed. Then, following the arguments of Theorem 1 It is easy to see that the topologies τ Conversely, we observe that D i is discontinuous on (F ∞ ⊕ F ∞ , τ ′ ∂ j ). For this, let ((a n , x n )) be a sequence in (F ∞ ⊕ F ∞ , τ ′ ∂ j ) such that it converges to (0, 0) in F ∞ ⊕ F ∞ . So, q ′ k,∂ j ((a n , x n )) → 0. Now we see that D i ((a n , x n )) = (0, ∂ i (a n )) does not converge to (0, 0). Since q ′ k,∂ j ((0, ∂ i (a n )) = p ′ k, j (∂ i (a n )), we show that p ′ k, j (∂ i (a n )) does not converge to 0. For this, suppose that for i = j, the Fréchet algebra topologies τ ′ ∂ j and τ ′ ∂ i are not equivalent. Then, for all n ∈ I N and c > 0, there exists some m ∈ I N such that Since q ′ n, ∂ i ((0, ∂ i (a))) = c p ′ n, i (∂ i (a)) and ∂ i is discontinuous on (F ∞ , τ i ), we see that p ′ k, j (∂ i (a n )) does not converge to 0 whenever a n converges to 0. So . Hence the result follows. ✷ We remark that for each i ∈ I Z + , the Fréchet algebra topologies τ ∂ i and τ ′ ∂ i are mutually inequivalent on the algebra F ∞ ⊕ F ∞ , because if they are equivalent, then, by Corollary 3.3 and by the transitivity property of the relation "equivalence of topologies", we see that for i = j, the Fréchet algebra topologies τ ∂ i and τ ′ ∂ j are equivalent on the algebra F ∞ ⊕ F ∞ , and so, for i, j ∈ I Z + with i = j, the Fréchet algebra topologies τ ′ ∂ i and τ ′ ∂ j are equivalent on the algebra F ∞ ⊕ F ∞ , a contradiction.
We give the following extension of Theorem 3.1 in the Fréchet case. The author believes that the answer of this question is in the affirmative, but has been unable to settle the question in general. For example, in the Banach case, one may modify the Feldman's example to have an example of a Banach algebra with uncountably many inequivalent complete algebra norms