{"id":5245,"date":"2012-11-21T19:42:19","date_gmt":"2012-11-21T18:42:19","guid":{"rendered":"http:\/\/djalil.chafai.net\/blog\/?p=5245"},"modified":"2023-02-12T16:03:16","modified_gmt":"2023-02-12T15:03:16","slug":"about-wigner-theorem","status":"publish","type":"post","link":"https:\/\/djalil.chafai.net\/blog\/2012\/11\/21\/about-wigner-theorem\/","title":{"rendered":"About the Wigner theorem"},"content":{"rendered":"

\"Matrix\"<\/a><\/p>\n

Recall that a random \\( {n\\times n} \\) Hermitian matrix \\( {M} \\) belongs to the Gaussian Unitary Ensemble (GUE) when it admits a density proportional to<\/p>\n

\\[ M\\mapsto \\exp\\left(-\\frac{n}{2}\\mathrm{Tr}(M^2)\\right). \\]<\/p>\n

If \\( {\\lambda_1,\\ldots,\\lambda_n} \\) are the eigenvalues of \\( {M} \\) then the Wigner theorem gives<\/p>\n

\\[ \\mathbb{E}\\left(\\frac{\\mathrm{Card}\\{1\\leq i\\leq n:\\lambda_i\\in I\\}}{n}\\right) \\underset{n\\rightarrow\\infty}{\\longrightarrow} \\mu_{\\mathrm{SC}}(I) \\]<\/p>\n

for every interval \\( {I\\subset\\mathbb{R}} \\), where \\( {\\mu_{\\mathrm{SC}}} \\) is the semicircle distribution with density \\( {\\frac{1}{2\\pi}\\sqrt{4-x^2}\\mathbf{1}_{[-2,2]}(x)} \\). In other words, the averaged empirical spectral distribution<\/p>\n

\\[ \\mu_n :=\\mathbb{E}\\left(\\frac{1}{n}\\sum_{i=1}^n\\delta_{\\lambda_i}\\right) \\]<\/p>\n

tends weakly to \\( {\\mu_{\\mathrm{SC}}} \\) as \\( {n\\rightarrow\\infty} \\). There are many proofs of this result. We present in this post a simple analytic proof due to Pastur.<\/p>\n

Cauchy-Stieltjes transform.<\/b> Set \\( {\\mathbb{C}_+=\\{z\\in\\mathbb{C}:\\mathfrak{I}z>0\\}} \\). The Cauchy-Stieltjes transform \\( {s_\\mu:\\mathbb{C}_+\\rightarrow\\mathbb{C}} \\) of a probability measure \\( {\\mu} \\) on \\( {\\mathbb{R}} \\) is given by<\/p>\n

\\[ s_\\mu(z):=\\int\\!\\frac{1}{\\lambda-z}\\,d\\mu(\\lambda). \\]<\/p>\n

The knowledge of \\( {s_\\mu} \\) on \\( {\\mathbb{C}_+} \\) (and even on an small ball of \\( {\\mathbb{C}_+} \\) by analyticity) is enough to characterize \\( {\\mu} \\). We always have the universal bound for any \\( {z\\in\\mathbb{C}_+} \\):<\/p>\n

\\[ |s_\\mu(z)|\\leq\\frac{1}{\\mathfrak{I}z}. \\]<\/p>\n

Furthermore, if \\( {{(\\mu_n)}} \\) is a sequence of probability measures on \\( {\\mathbb{R}} \\) such that \\( {s_{\\mu_n}(z)\\rightarrow s(z)} \\) as \\( {n\\rightarrow\\infty} \\) for every \\( {z\\in\\mathbb{C}_+} \\) then there exists a unique probability measure \\( {\\mu} \\) on \\( {\\mathbb{R}} \\) such that \\( {s_\\mu=s} \\) and such that \\( {\\mu_n\\rightarrow\\mu} \\) weakly as \\( {n\\rightarrow\\infty} \\).<\/p>\n

Our problem reduces to show that \\( {s_{\\mu_n}(z)\\rightarrow s_{\\mu_{\\mathrm{SC}}}(z)} \\) for every \\( {z\\in\\mathbb{C}_+} \\).<\/p>\n

Resolvent.<\/b> The resolvent \\( {G} \\) of a Hermitian matrix \\( {H} \\) at point \\( {z\\in\\mathbb{C}_+} \\) is<\/p>\n

\\[ G(z)=(H-zI)^{-1}. \\]<\/p>\n

The map \\( {z\\mapsto G(z)} \\) is holomorphic on \\( {\\mathbb{C}_+} \\) and<\/p>\n

\\[ \\left\\Vert G(z)\\right\\Vert_{2\\rightarrow2}\\leq \\frac{1}{\\mathfrak{I}z}. \\]<\/p>\n

Denoting \\( {\\tau:=\\frac{1}{n}\\mathrm{Tr}} \\) and \\( {\\mu:=\\frac{1}{n}\\sum_{k=1}^n\\delta_{\\lambda_k(H)}} \\), we have<\/p>\n

\\[ s_\\mu(z)=\\tau(G(z)). \\]<\/p>\n

The resolvent identity<\/em> states that if \\( {G_1} \\) and \\( {G_2} \\) are the resolvent at point \\( {z} \\) of two Hermitian matrices \\( {H_1} \\) and \\( {H_2} \\) then<\/p>\n

\\[ G_2-G_1=-G_1(H_2-H_1)G_2. \\]<\/p>\n

Used for \\( {H_1=M} \\) and \\( {H_2=0} \\), this gives, taking expectations and multiplying both sides by \\( {z} \\), and denoting \\( {G} \\) the resolvent at point \\( {z} \\) of \\( {M} \\),<\/p>\n

\\[ -I-z\\mathbb{E}(G)=-\\mathbb{E}(GM). \\]<\/p>\n

From this identity, we will obtain an equation satisfied by \\( {s_{\\mu_n}=\\mathbb{E}(\\tau(M))} \\).<\/p>\n

Integration by parts.<\/b> For a random vector \\( {X} \\) with density \\( {e^{-V}} \\) on \\( {\\mathbb{R}^d} \\) and a smooth \\( {F:\\mathbb{R}^d\\rightarrow\\mathbb{C}} \\), an integration by parts gives<\/p>\n

\\[ \\mathbb{E}(F(X)\\nabla V(X))=\\mathbb{E}(\\nabla F(X)) \\]<\/p>\n

as soon as \\( {e^{-V(x)}F(x)\\rightarrow0} \\) as \\( {\\left\\Vert x\\right\\Vert\\rightarrow\\infty} \\). In particular, for every \\( {1\\leq p,q\\leq n} \\), by denoting \\( {\\partial_{M_{pq}}:=\\frac{1}{2}(\\partial_{\\Re M_{pq}}-i\\partial_{\\mathfrak{I} M_{pq}})} \\) if \\( {p\\neq q} \\),<\/p>\n

\\[ \\mathbb{E}(F(M)M_{pq}) =\\frac{1}{n}\\mathbb{E}(\\partial_{M_{qp}}F(M)). \\]<\/p>\n

Let us examine the case \\( {F(M)=G_{ab}} \\) with \\( {1\\leq a,b\\leq n} \\). The resolvent identity used twice gives<\/p>\n

\\[ \\begin{array}{rcl} G_2-G_1 &=&-G_1(M_2-M_1)G_2 \\\\ &=&-G_1(M_2-M_1)(G_1-G_1(M_2-M_1)G_2)\\\\ &=&-G_1(M_2-M_1)G_1+o(\\left\\Vert M_2-M_1\\right\\Vert) \\end{array} \\]<\/p>\n

which gives in particular that for every \\( {1\\leq a,b,p,q\\leq n} \\),<\/p>\n

\\[ \\partial_{M_{pq}}G_{ab}=-G_{ap}G_{qb} \\]<\/p>\n

As a consequence, we get, using the integration by parts, for every \\( {1\\leq p\\leq n} \\),<\/p>\n

\\[ \\mathbb{E}(GM)_{pp} =\\sum_{q=1}^n\\mathbb{E}(G_{pq}M_{qp}) =\\frac{1}{n}\\sum_{q=1}^n\\mathbb{E}(\\partial_{pq}G_{pq}) =-\\frac{1}{n}\\sum_{q=1}^n\\mathbb{E}(G_{pp}G_{qq}) =-\\mathbb{E}(G_{pp}\\tau(G)). \\]<\/p>\n

Summing up, we obtain, after taking normalized traces,<\/p>\n

\\[ -1-zs_{\\mu_n}(z)=\\mathbb{E}(\\tau(G)^2). \\]<\/p>\n

Poincar\u00e9 inequality.<\/b> To compare \\( {\\mathbb{E}(\\tau(G)^2)} \\) in the right hand side with \\( {\\mathbb{E}(\\tau(G))^2=s_{\\mu_n}(z)^2} \\), we may use concentration<\/a>, or alternatively and following Pastur, a Poincar\u00e9 inequality. Namely, we set \\( {f(M):=\\tau(G)} \\) and we write<\/p>\n

\\[ \\left|\\mathbb{E}(\\tau(G)^2)-(\\mathbb{E}\\tau(G))^2\\right| = \\left|\\mathbb{E}((\\tau(G)-\\mathbb{E}\\tau(G))^2)\\right| \\leq \\mathbb{E}(\\left|\\tau(G)-\\mathbb{E}\\tau(G)\\right|^2) =\\mathrm{Var}(f(M)). \\]<\/p>\n

Now \\( {M} \\) is Gaussian with covariance matrix \\( {C} \\) such that \\( {\\left\\Vert C\\right\\Vert_{2\\rightarrow2}\\leq \\mathcal{O}(n^{-1})} \\). Thus, the Poincar\u00e9 inequality for the law of \\( {M} \\) and for function \\( {f} \\) gives<\/p>\n

\\[ \\mathrm{Var}(f(M)) \\leq \\mathcal{O}(n^{-1})\\mathbb{E}(\\left\\Vert \\nabla f(M)\\right\\Vert_2^2). \\]<\/p>\n

But<\/p>\n

\\[ \\partial_{M_{pq}}\\tau(G) =\\frac{1}{n}\\sum_a\\partial_{M_{pq}}G_{aa} =-\\frac{1}{n}\\sum_aG_{ap}G_{qa} =-\\frac{1}{n}G^2_{qp} \\]<\/p>\n

and therefore<\/p>\n

\\[ \\left\\Vert \\nabla f\\right\\Vert_2^2 =\\frac{1}{n^2}\\sum_{p,q}\\left|\\partial_{M_{pq}}\\tau(G)\\right|^2 =\\frac{1}{n^2}\\left\\Vert G^2\\right\\Vert_{\\mathrm{HS}}^2 \\leq \\frac{1}{n}\\left\\Vert G^2\\right\\Vert_{2\\rightarrow2}^2 \\leq \\frac{1}{n}\\left\\Vert G\\right\\Vert_{2\\rightarrow2}^4 \\leq \\frac{1}{n(\\mathfrak{I} z)^4}. \\]<\/p>\n

Putting all together, this gives finally<\/p>\n

\\[ \\left|s_{\\mu_n}(z)^2+zs_{\\mu_n}(z)+1\\right| = \\mathcal{O}\\left(\\frac{1}{n^2(\\mathfrak{I} z)^4}\\right) \\underset{n\\rightarrow\\infty}{\\longrightarrow}0. \\]<\/p>\n

Conclusion.<\/b> The equation \\( {s^2+zs+1=0} \\) in \\( {s} \\) has two roots \\( {s_\\pm(z):=\\frac{1}{2}(-z\\pm\\sqrt{z^2-4})} \\). We observe that \\( {\\mathfrak{I}s_{\\mu}(z)\\mathfrak{I}z>0} \\). This gives that \\( {\\lim_{n\\rightarrow\\infty}s_n(z)=\\frac{1}{2}(-z+\\sqrt{z^2-4})} \\) for every \\( {z\\in\\mathbb{C}_+} \\), which is the Cauchy-Stieltjes transform of \\( {\\mu_{\\mathrm{SC}}} \\). We conclude that \\( {\\mu_n} \\) converges weakly as \\( {n\\rightarrow\\infty} \\) to \\( {\\mu_{\\mathrm{SC}}} \\). One may drop the expectation in \\( {\\mu_n} \\) and obtain an almost sure weak convergence by using concentration<\/a> and the first Borel-Cantelli lemma.<\/p>\n

Further reading.<\/b> This method is quite powerful and can be extended to many models of random matrices, including Wigner matrices (not necessarily Gaussian), as explained in the book by Pastur and Shcherbina<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"

Recall that a random \\( {n\\times n} \\) Hermitian matrix \\( {M} \\) belongs to the Gaussian Unitary Ensemble (GUE) when it admits a density…<\/p>\n

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