If two projections are close, then they are unitarily equivalent

Given two projections $p,q\in B(H)$, it is well-known that if $\|p-q\|<1$, then there exists a unitary $u\in B(H)$ with $q=upu^*$.

The proof that immediately occurs to me uses comparison of projections (the inequality forces $p\sim q$ and also $p^\perp\sim q^\perp$, and then one constructs the unitary by adding the corresponding partial isometries), and implies that one can even replace $B(H)$ with any von Neumann algebra that contains both $p$ and $q$.

Questions:

1. Does anyone know of a more direct proof? Comparison of projections is a very basic result in von Neumann algebras, but somehow here it feels like overkill.
2. Does this property hold in a C$^*$-algebra? Proof, or counterexample?
• Here is a ring proof. Let $R$ be a ring with $1$. If both $a,b$ are idempotents in $R$ st $1\pm (a-b)$ are invertible, then there is a simple argument to show that $a = xbx^{-1}$ for some invertible $x$ in $R$. In a C*-algebra ordinary equivalence implies *-equivalence, and you're done. – David Handelman May 14 '18 at 2:30

The result holds in a C*-algebra. The sketch of the proof I know, which was given to me as a series of exercises in a class on C*-algebras, is as follows. First one constructs isomorphisms from each C*-algebra on the following list to the next one (below all C*-algebras are unital, $\mathbb{Z}_2$ denotes the cyclic group of order $2$, and $M_2$ denotes the C*-algebra of $2 \times 2$ complex matrices):

• The free C*-algebra on two projections $p, q$,
• The free C*-algebra on two self-adjoint unitary elements,
• The group C*-algebra $C^{\ast}(\mathbb{Z}_2 \ast \mathbb{Z}_2)$,
• The crossed product $C(\mathbb{T}) \rtimes_{\alpha} \mathbb{Z}_2$ where $\mathbb{Z}_2$ acts on $\mathbb{T}$ by complex conjugation,
• The C*-algebra $A$ of continuous functions $g : [0, 1] \to M_2$ such that $g(0)$ and $g(1)$ both have the form $\left[ \begin{array}{cc} \alpha & \beta \\ \beta & \alpha \end{array} \right]$ for some $\alpha, \beta$ (possibly different for $0$ and $1$).

Under one such chain of isomorphisms $p$ gets sent to the function

$$[0, 1] \ni x \mapsto \left[ \begin{array}{cc} \frac{1}{2} & \frac{e^{\pi i x}}{2} \\ \frac{e^{-\pi i x}}{2} & \frac{1}{2} \end{array} \right]$$

and $q$ gets sent to the function with constant value $\left[ \begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{array} \right]$.

The imposition of the additional constraint $\| p - q \| = \lambda < 1$ corresponds to working in a quotient of $A$ where we consider functions $g : [0, x_0] \to M_2$ where $\sin \frac{\pi x_0}{2} = \lambda < 1$ (this condition is precisely what is needed for the sup norm of $p - q$, as a function to $M_2$, to be equal to $\lambda$). Then it suffices to write down a unitary equivalence between $p$ and $q$ in this algebra, but it becomes straightforward to do this explicitly: in fact we can take

$$u : x \mapsto \left[ \begin{array}{cc} 1 & 0 \\ 0 & e^{\pi i x} \end{array} \right].$$

(Note that $u$ does not define an element of $A$.)

Edit: I'm leaving the old post below, but before I want to write the proof as suggested by Bruce from his book, which uses the ideas in a more efficient way.

Assume that $\|p-q\|<1$, with $p,q\in A$, a unital C$^*$-algebra. Let $x=pq+(1-p)(1-q)$. Then, as $2p-1$ is a unitary, $$\|1-x\|=\|(2p-1)(p-q)\|=\|p-q\|<1.$$ So $x$ is invertible. Now let $x=uz$ be the polar decomposition, $z=(x^*x)^{1/2}\in A$. Then $u=xz^{-1}\in A$. Also, $px=pq=xq$, and $qx^*x=qpq$, so $qx^*x=x^*xq$, and then $qz=zq$. Then $$pu=pxz^{-1}=xqz^{-1}=uzqz^{-1}=uqzz^{-1}=uq.$$ So $q=u^*pu$.

============================================= (the old post starts here)

(A good friend pointed me to the ideas in this answer, so I'm sharing them here)

The result holds in any unital C$^*$-algebra. So assume that $\|p-q\|<1$, with $p,q$ in a unital C$^*$-algebra $A\subset B(H)$.

Claim 1: There is a continuous path of projections joining $p$ and $q$.

Proof. Let $\delta\in(0,1)$ with $\|p-q\|<\delta$. For each $t\in[0,1]$, let $x_t=tp+(1-t)q$. Then $$\|x_t-p\|=\|(1-t)(p-q)\|<\delta(1-t),$$ $$\|x_t-q\|=\|t(p-q)\|<\delta t.$$ This, together with the fact that $x_t$ is selfadjoint, implies that $\sigma(x_t)\subset K=[-\delta/2,\delta/2]\cup[1-\delta/2,1+\delta/2]$ (since $\min\{t,1-t\}\leq1/2$). Now let $f$ be the continuous function on $K$ defined as $0$ on $[-\delta/2,\delta/2]$ and $1$ on $[1-\delta/2,1+\delta/2]$. Then, for all $t\in[0,1]$, $f(x_t)\in A$ is a projection. And $$t\to x_t\to f(x_t)$$ is continuous, completing the proof of the claim. Edit: years later, I posted this answer to a question on MSE that proves the continuity.

Claim 2: We may assume without loss of generality that $\|p-q\|<1/2$.

This is simply a compacity argument, using that each projection in the path $f(x_t)$ is very near another projection in the path. The compacity allows us to make the number of steps finite, and so if we find projectons $p=p_0,p_1,\ldots,p_n=q$ and unitaries with $u_kp_ku_k^*=p_{k+1}$, we can multiply the unitaries to get the unitary that achieves $q=upu^*$.

Claim 3: If $\|p-q\|<1/2$, there exists a unitary $u\in A$ with $q=upu^*$.

Let $x=pq+(1-p)(1-q)$. Then $$\|x-1\|=\|2pq-p-q\|=\|p(q-p)+(p-q)q\|\leq2\|p-q\|<1,$$ so $x$ is invertible. Let $x=uz$ be the polar decomposition. Then $u$ is a unitary. Note that $$qx^*x=q(qpq+(1-q)(1-p)(1-q))=qpq,$$ so $q$ commutes with $x^*x$ and then with $z=(x^*x)^{1/2}$. Note also that $px=xq$, so $puz=uzq=uqz$. As $z$ is invertible, $pu=uq$, i.e. $$q=u^*pu.$$ Note that $u=xz^{-1}\in A$.

• The first part of the argument can be done without, since the element $x$ is invertible from the hypotheses. Say $\|p-q\|=\delta<1$ ( so that $q\geq p-\delta$ and also $1-q\geq (1-p)-\delta$). Then $x^*x=pqp+(1-p)(1-q)(1-p)\geq p(p-\delta)p+(1-p)(1-p-\delta)(1-p)=(1-\delta)$. So $x^*x$ is invertible and similarly $xx^*$ is invertible. – Leonel Robert Dec 11 '13 at 15:15
• In fact, this proof works if $\|p-q\|<1$ and gives $u\in A$; in the example in the remark $\|p-q\|=1$ and $p$ and $q$ are not equivalent in $A$. Additionally, $u$ is a continuous function of $p$ and $q$. See Proposition II.3.3.4 of my Operator Algebras book for this (folklore) result. – Bruce Blackadar Dec 11 '13 at 15:19
• Thanks, Bruce! I should have looked at your book first, but it was kind of easier to ask here :D – Martin Argerami Dec 11 '13 at 21:31
• The revised proof you wrote is not quite complete, since $x^*x$ invertible does not imply $x$ invertible (you also need $xx^*$ invertible). But in fact in this case $1-x^*x=1-xx^*=(p-q)^2$, so $\|1-x^*x\|=\|1-xx^*\|<1$ and both $x^*x$ and $xx^*$ are invertible. Actually there is an even simpler argument: $1-x=(2q-1)(q-p)$, so $\|1-x\|=\|q-p\|<1$ since $2q-1$ is unitary. – Bruce Blackadar Dec 11 '13 at 21:33
• Sorry, I got $p$ and $q$ interchanged in the last comment. – Bruce Blackadar Dec 11 '13 at 21:49

In Two Subspaces (Trans. AMS 1969), Halmos made some useful observations about pairs of subspaces/projections. When $\|p-q\|<1$, this yields in particular that we can assume, up to unitary equivalence, that $$p=\pmatrix{1&0\\0&0} \qquad q=\pmatrix{c^2& cs\\cs& s^2}\qquad c^2+s^2=1$$ where $c, s$ are commuting positive contractions which allow computations just like for pairs of rank one projections in the two-dimensional case where $c=\cos \theta$ and $s=\sin\theta$. Then it is natural to come up with the following candidate for the unitary you want, namely $$u=\pmatrix{c & s\\s&-c}$$ Once we have found that, we just have to express this in terms of $p$ and $q$. And we find

$$u=(p+q-1)|p+q-1|^{-1}$$ which is a self-adjoint involution, lying obviously in the $C^*$-algebra geberated by $p$ and $q$.

Note To check that the above formula does the job, the key point is that $(p+q-1)^2=1-(p-q)^2$ is invertible, positive, and commutes with $p$ and $q$.

Remark Another candidate would be the "rotation" $$v=\pmatrix{c& -s\\ s& c}\qquad \mbox{i.e.} \quad v=u(2p-1)$$ An advantage of the latter unitary is that we have $$\|v-1\|\leq \sqrt{2}\|p-q\|$$ If I recall correctly, this is mentioned in Takesaki I for reference purposes.

• Nice. In contrast to other proofs, the $u$ you have given yields an automorphism of $A$ which exchanges $p$ and $q$ (rather than just taking one to the other) which is exactly what I came here looking for. It also has the continuity and functorial properties of other constructions, so I wonder why this isn't standard? – Tristan Bice Nov 22 '14 at 19:40

Qiaochu give a "real" proof that works in any C*-algebra. If all you care about is $B(H)$, we can give a very simple proof because all you need for $p$ and $q$ to be unitarily equivalent is that their ranges have the same dimension, and their kernels have the same dimension. Well, if two closed subspaces have different dimensions then the one with larger dimension intersects the orthocomplement of the other [proof below], and this easily implies that $\|p - q\| < 1$ forces $p$ and $q$ to have ranges and kernels with matching dimensions.

[Proof of the claim: first suppose $E$ and $F$ are subspaces of a finite dimensional space and ${\rm dim}(E) < {\rm dim}(F)$. Then $n < {\rm dim}(E^\perp) + {\rm dim}(F) = {\rm dim}(E^\perp \vee F) + {\rm dim}(E^\perp \wedge F)$ where $n$ is the dimension of the ambient space. So ${\rm dim}(E^\perp \wedge F) > 0$. Now if $E$ and $F$ are subspaces of a separable infinite dimensional space and ${\rm dim}(E) < {\rm dim}(F)$ then ${\rm dim}(E)$ is finite and we can find a finite dimensional subspace $F_1$ of $F$ with ${\rm dim}(F_1) < {\rm dim}(E)$. Then run the preceding argument within the finite dimensional space $E \vee F_1$.]

I know of a slightly more direct proof of the original problem.

1. If p and q are projections in B(H) then p ~ q if and only if the dimensions of range of p and the range of q are equal.

2. If || p-q|| < 1, then dimensions of range of p and range of q are equal.

Then one can proceed as you mentioned in the question.

• Not quite --- you also need the dimensions of their kernels to be equal. – Nik Weaver Dec 11 '13 at 2:41
• Well- kernel of p= range of 1-p. ||(1-p)- (1-q)||=||p-q||<1. – voldemort Dec 11 '13 at 3:48